Transcript
INTEGRAL
CALCULUS
FOR
BEGINNERS
INTEGRAL
FOR
CALCULUS
BEGINNERS
WITH
AN
INTRODUCTION
TO
THE
STUDY
OF
DIFFERENTIAL
EQUATIONS
BY
JOSEPH
FORMERLY
FELLOW OF
EDWAEDS,
SIDNEY
SUSSEX
M.A.
COLLEGE,
CAMBRIDGE
MACMILLAN
AND NEW
AND
YORK
CO.
1896
All
rightsreserved
First
Edition, 1891, corrections,
1890.
Reprinted
With additions
1892,
1894
1893.
and
;
reprinted
1896.
GLASGOW
:
PRINTED
AT
THE
UNIVERSITY
PRESS
BY
ROBERT
MACLEHOSE
AND
CO.
PREFACE.
THE
present
volume
to
a
is
intended of the
the
to
form
a
sound
introduction suitable for
a
study
Integral Calculus, subject.
Like its
student
beginning
companion,
it does
at
are
the
Calculus 'Differential! aim all
as
for Beginners,
but rather which
not
therefore of
at
completeness,
of the for the
a
the
omission
portions
best
subject
later
usually regarded
It will of be
left that
reading.
cesses prothe
found,
however,
are
ordinary
as
integration
methods calculation of
to the
fully treated,
and
also
principal
and of the solids
of
Rectification of
the
Quadrature,
and surfaces
is
volumes Some
revolution. student of
indication useful
the the
also
afforded of
to the
other
as
applications
method of
a
Integral Calculus, employed
or
such
general
be
in
obtaining
of
a
position
of Inertia. of
a
Centroid,
As
in want
the
value
Moment
the
it
seems
undesirable Mathematics with
that should
the
path
be
student
Applied
of
blocked of
by
a
acquaintance
methods
solving
M298720
vi
PREFACE.
elementaryDifferential Equations,and
time that his
course
at the
a
same
should the
be
stopped for
some
tematic sys-
study
and added
more
of
subject in
brief of
complete
has been of the
to and
exhaustive of
the
treatise,a
account
ordinarymethods
such
solution
forms occurring, elementary leading up kinds
as
all including
meet
the of
student
is
likelyto
with of
in
a
his
reading
to the
Analytical Statics,
Dynamics
Linear
and Particle,
the
elementary parts of general
cients, Coeffiis
Rigid Dynamics. Up
Differential the consistent The been
solution of the with treated
Equation
been
Constant
as
subjecthas
with
fullyas
text
the scope
of the
present work.
have
examples scattered throughoutthe
made carefully
or
selected
to
illustrate the A be able considerworked
articles which number
follow. they immediately of
so
these
examples should
be
by
in
the student the
"
that the several methods
"
explained
in sets
book-work
may
firmly fixed
harder
the
at
a
mind
the
more
before
ends
the attacking
somewhat
are
of the
chapters.These
character, and
generallyof
call for
miscellaneous
greater
these
and originality considerable
ingenuity, though
been set actually in
few
present any
A largeproportion of difficulty.
examples have
the
sources
examinations, and
for them
are
to
which
I
am
indebted
usuallyindicated.
PREFACE.
vii
My
the
acknowledgments
works of
many
are
due
in
some
degree
on
to
of
the
modern
writers
the
subjects
Treatises
treated
of,
Bertrand
but
more
especially
and
to
the
of
and
Todhunter,
to
fessor Pro-
Greenhill's
interesting
the
more
Chapter
advanced
on
the
Integral
may
Calculus,
consult
with
which
student
great
are
advantage.
due
to
My kindly
the
thanks
several
friends
who
have
sent
me
valuable
suggestions
with
regard
to
desirable
scope
and
plan
of
the
work.
JOSEPH
EDWARDS.
October,
1894.
CONTENTS.
INTEGRAL
CALCULUS.
CHAPTER
I.
NOTATION,
Determination of from
SUMMATION,
APPLICATIONS.
PAGES
an
Area,
......
1
"
3
Integration
Volume of
the
Definition,
10
4
"
9 13
Revolution,
"
CHAPTER
II.
GENERAL
METHOD.
STANDARD
FORMS.
Fundamental Nomenclature General Laws of
Theorem,
.......
14
"
19
21
"
and
Notation,
......
20
obeyed
xn,
by
the
Integrating
Symbol,
. .
22 23"26 26"28
Integration
Table of
x~l,
Results,
CHAPTER
III.
METHOD
OF
SUBSTITUTION.
Method The
of
Changing
the
Variable,
......
29
"
32 36
"
Hyperbolic
Functions,
33
Additional
Standard
Results,
37"41
CONTENTS.
CHAPTER INTEGRATION
BY
IV. PARTS.
PAGES
"by Integration
Geometrical Extension of the
Parts"
of
a
Product,
43"47
....
Proof, Rule,
48"49 50"52
CHAPTER PARTIAL Standard General
V. FRACTIONS. 55
"
Cases,
Fraction
57
........
with
Rational
Numerator
and
nominator, De58"61
CHAPTER
SUNDRY
VI. METHODS.
STANDARD
Integrationof
Powers Powers Powers and
f^L
J \/K
or
65"68
... .
Products
of Sines and
Cosines,
.....
.
.
.
69"74
75
"
of Secants of
Cosecants,
or
76
78
Tangents
a
Cotangents,
"
77
.....
"
/rfv
+ o
cos
,
etc.,
79"83
x
CHAPTER REDUCTION
of xm-lXP, Integration Reduction Formulae where X
VII. FORMULAE.
=
a
+
bxn,
....
.
.
.
87"89
90
"
for
/ xm~lXpdx,
93
Reduction
Formulae
7T
for
/ sin^a; cos^a: dx,
IT
.
.
.
94
"
95
Evaluation
of
/ sii\nxdx,
j"z
r -i
I
ain^x cos^x
dx,
.
.
.
96
"
102
CONTENTS.
xi
CHAPTER MISCELLANEOUS
VIII. METHODS.
PAGES
Integrationof
J
/^).fx.9 X. Y
.......
109"117
118
.
i\f
of some Integration SpecialFractional Forms, General and Geometrical Illustrations, Propositions Some Elementary Definite Integrals, Differentiation under an Integral Sign,
. .... ....
"
119
"
120
.
124
125 128
"
127
129
"
CHAPTER RECTIFICATION. Rules for
IX.
Curve-Tracing,
for Rectification for
a
.......
135"137
Illustrative
.....
Formulae
and
Examples,
.
13S
"
139 140 143
Modification Arc of
an
Closed Curve,
........
Evolute,
Intrinsic Arc
Equation,
Curve,
........
144
"
149 150
of Pedal
........
CHAPTER
X.
QUADRATURE.
Cartesian
Formula,
Closed
........
153
.......
"
157
Sectorial Areas. Area Other Area of
a
Polars, Curve,
.....
158"160 161"163
.......
Expressions,
between
a
'.''-.
.
164"165
Curve,
.........
two
Radii
of Curvature
and
the
Evolute,
Areas of
166"167
........
Pedals,
168"175
.......
CorrespondingAreas,
176
"
177
CHAPTER SURFACES Volumes of
AND
XI.
OF
VOLUMES
SOLIDS
OF
REVOLUTION. 183"184
Revolution, Surfaces of Revolution,
.......
.......
185"187
xii
CONTENTS.
PAGES
Theorems Revolution
of
Pappus,
a
.......
188
......
191
of
Sectorial
Area,
192
CHAPTER
XII.
OF
SECOND-ORDER
ELEMENTS
AREA.
MISCELLANEOUS
APPLICATIONS. Surface Cartesian Integrals,
; Moments
Element,
.....
.
195 199
"
198
201
Centroids Surface
of
Inertia,
"
Polar Element, Integrals, etc.,Polar Formulae, Centroids,
202"203
.....
204
"
207
DIFFERENTIAL
CHAPTER
EQUATIONS.
XIII.
THE
EQUATIONS
Genesis Variables Linear of
a
OF
FIRST
ORDER. 211"214 215 216
"
Differential
Equation,
.....
Separable, Equations,
219
CHAPTER
XIV. ORDER
EQUATIONS
OF
THE
FIRST
(Continued}.
221
"
Homogeneous Equations,
One Letter
226
Absent, Form,
227"229
230"233
Clairaut's
CHAPTER
XV. ORDER. EXACT DIFFERENTIAL
EQUATIONS
OF
THE
SECOND
EQUATIONS.
Linear One
Equations,
Absent,
of
.
235
"
236
Letter
237"238
a
General Exact
Equation. Removal Differential Equations,
Linear
.
Term,
. .
239
. . . .
"
240
241"242
CONTENTS.
xiii
CHAPTER LINEAR DIFFERENTIAL
XVI.
EQUATION
COEFFICIENTS.
WITH
CONSTANT
PAGES
General The The
An
Form
of
Solution, Function,
.....
243"244 245
"
Complementary
Particular
251
Integral,
Reducible
to Linear
252"263 Form with Constant 264"265
Equation
Coefficients,
CHAPTER ORTHOGONAL TRAJECTORIES.
XVII. MISCELLANEOUS
EQUATIONS.
266"269
Orthogonal
Some Further
Trajectories, Dynamical Equations,
....
Important
270
"
271
Illustrative
Examples,
272"277
Answers,
278"308
ABBREVIATION. To indicate
cases
the
sources
from
a
which of
many have
as
of
the
an
examples
examination
"
are
derived, in
in
common,
where
group
are
colleges
held follows
the St.
references
abbreviated
:
(a)
=
Peter's, Pembroke,
Corpus
Christi, Queen's,
and
St.
Catharine's.
(j8)
=
Clare, Caius, Trinity Hall, and Jesus, Christ's, Magdalen,
Jesus,
King's.
and
(7)
(d)
=
Emanuel,
and
Sidney
Sussex.
=
Christ's, Emanuel,
Sidney
Sussex.
(e)
=
Clare, Caius, and
King's.
INTEGRAL
CALCULUS
CHAPTER
I.
NOTATION,
SUMMATION,
APPLICATIONS.
1. The
Use
and
Aim
of
the is
Integral
the
Calculus.
of of
an deavour en-
Integral
to
Calculus
some
outcome
obtain
general
space bounded
method
finding
the
area
of
the
plane
by
given
of
curved
lines. In
area
the it is
a
problem
necessary
of
to
the
determination this of
some
such
an
suppose
space small of
divided elements.
up
into We the
each
very
large
have of
to
number form
sum
very
then limit is
method all these
obtaining
when
the
of
elements small and
ultimately
infinitesimally
increased. that when it
may
once
their
number It will
infinitely
be found is such
areas
such be
a
method
to
a
of other
summation
discovered,
as
applied
length
of
problems
line,
volumes
moments
E. i. c.
the of
finding
surfaces
of of
the
curved the of
etc.
"E
the
given
the
shape
and
bounded of
by
the
them,
determination of
inertia,
positions
A
Centroids,
2
INTEGRAL
CALCULUS.
Throughout the book all coordinate all angles will supposed rectangular,
measured
in
axes
will be
circular
measure,
and
be supposed all logarithms stated.
supposed Napierian, exceptwhen
of 2. Determination Notation. be Summed.
an
otherwise
Area.
Form
of Series to
of the portion to find the area Supposeit is required of space bounded AB, defined by by a given curve of and BM its Cartesian equation, the ordinates AL A and
B, and the cc-axis.
L
0,0,0,0,
Fig. 1.
Let the
=
LM
Q^Qv QiQz,
be divided into n equal small "f lengthA, and let eacn """"
"
LQV parts,
a
Then b abscissae of A and ". a of the curve, be the equation (f)(x) y LA, QiPp $2^2*e^c-'through the several
=
and 6 be ?i/L Also if the ordinates
points L, of lengths are etc., ^(a+K), ^(a+2A),etc. "j"(a), Qv Q2, Let their extremities be respectively A, P1? P2, etc., the rectangles and complete AQV PjQg,P2Q3,etc. of these n rectangles falls short of Now the sum of the n small figures, the area sought by the sum etc. Let each of these be supposed 1, P1J22P2,
INTEGRAL
CALCULUS.
the
the may
term
h(f"(a+nh) or
is taken.
A0(6)
which limit
vanishes of this
when series
limit
Hence
the
also be written f6
I
a
"t"(x)dx.
3.
Integration from
summation
as
the
Definition.
be effected
:
"
This
may
we
sometimes
now
by
elementary means,
Ex.
1.
proceed
to illustrate
Cb
Calculate
/ e*dx.
Here
we
have
to
evaluate
+ ea+"
+ ea+h Lth==Qh[ea
+
.
.
.
where
This
b
=
a
+ nh.
ea)-^-=e* =Lth^h^p\ea=Lth^(eb
-
e\
"
1
"
"
X
[By Diff. Calc. for Beginners,
Art.
15.]
/b
Now
r=n-l
xdx
we
have
to find
Lt
2
r="
("+rA)A,
where
2(a + rh)h
=
and
in the
limit
becomes
2
22'
obtain the limit when h is
/61 "$x
a
we
have
to
diminished indefinitely
of
NOTATION,
SUMMATION,
APPLICATIONS.
5
"
a
b+ h
a-h and when
"'
without each limit, of these becomes
h diminishes
II
a
b'
Thus
J
a
/"==*.* f*JL
.r2
a
b
Ex. 4.
Prove
ab initio that
/"
We
now
are a [sin
sin
#
ofo?
=
cos
a
-
cos
6.
to find the limit of + + 2A)+ + k)+ sin(a sin(a l- Jsin n2/ 2,
. . .
to
n
terms]A,
sinf a+n
\
"
sin
|
*
"
-
This
expression
=
cosf a
J
"
cos
" a
+
(2n
-
1)-j"
-
2JJsin2
sm-
which
when
A is
small ultimately takes indefinitely
cos a
"
the form
cos
b.
INTEGRAL
CALCULUS.
EXAMPLES.
Prove
by
summation
that
2.
/ sinh sir xdx /
"
cosh b
"
cosh
a.
3.
/b
4. As
a
cos
OdO
=
sin 6
"
sin
a.
of Integration
further
xm.
example we
sum
next
propose
to consider
the limit of the
of the series
h[am+ (a+ h}m+ (a+ 2h)m+.
where
i
6
7 h
=
--
"
a
,
n
and
n
is made
The
m indefinitely large,
+ 1 not
1
being zero.
when A be is of
[Lemma.
"
Limit
of
fy v"/
I
I\m+l
_
%
2
yin +
"
-
is
m
+ 1
Aym
may
whatever diminished, indefinitely y finite magnitude. For the expression be written may
be, provided it
-1
y and less since h is to be than
zero ultimately we
may
consider the
-
to
be
y
unity,and
/
we
may
l
,
therefore whatever
apply
Binomial of m+l.
Theorem
to
expand
^\7?l+ ( 1 -J--
J
be the value
NOTATION,
SUMMATION,
APPLICATIONS.
7
(See Dif.
"becomes
Gale,
Art. for Beginners,
13.)
Thus
the
expression
-x(a convergent series)
y
"m
+ I when
A is
diminished.] indefinitely
In the result
put
and
i/ success! vely
=
a,
a+h, a+2h,etc....a + (n
(
"
l)h,
we
get
l-am+l_
~
T
r,
_
1
_
(a + n^
h(a+n-Ui)m
or
adding numerators
for
a new
for
a
new
numerator
and
nominat de-
denominator,
+ (a+ /t)w + (a + 2h)m+ fe[aw
or
.
.
.
+
(a+ n^l
+ (a+ A)m+ (a Lth=Qh[am
'
m+1
In may accordance be written
'6
with
the b"
notation
of Art.
2, this
xmdx="
7
m+1
8
The letters
INTEGRAL
CALCULUS.
whatever, represent any finite quantities become infinite between
x=a
and b may providedxm does not
a
a as
and
is taken is necessary in the
When
small exceedingly to proof suppose h an that
and ultimately zero, it infinitesimal of higher limit
-
order,for it has been assumed all the values givento y.
When 6
=
in the
is
zero
for
V
1 and
a
=
the 0, ultimately
theorem
comes be-
xmdx=
o
"
"7
if
m
+ 1 be
positive,
or
=
oo
if m
+ 1 be
negative.
This theorem
may
be written
also
"r
as according
m+1
is positive or
negative.The
limit
or,
which
is the
same
thing,
"M4-'
-Lstn=
oo
differs from and
the
former
1
"
by
-"?,
"
,
i.e. by 0 in the
oo
limit,
is will
n or as m+1 according
is therefore also
or positive negative. The
case
when
m
+ l=0
be discussed later.
Ex. Find the of the the of portion the parabola 7/2=4a# the ordinate x"c.
1.
area
bounded
by
the curve,
and #-axis,
NOTATION,
Let
us
SUMMATION,
APPLICATIONS.
divide
NM
is the
length c into n equal portionsof which Then if (r+l)th,and erect ordinates NP, MQ.
the
PR when
be
n
drawn
to NM, parallel
sum
the
is infinite of the
of such
required is the limit PM as (Art. 2), rectangles
area
i.e. where Now nh
=
Lt^PN.NM
c.
or
[By
Area
Art.
4.]
=f
extreme
=f
of the
area
of the Ex.
rectangleof which the are adjacentsides.
Find the
mass
ordinate
arid abscissa
2.
of
a
rod
whose
densityvaries
as
the
of the distance from one end. with power Let a be the length of the rod, o" its sectional area supposed uniform. Divide the rod into n elementary portions each of
length
of
1
\
"
-
-.
The densitv
volume is w-,
n mass
of the and
(r+l)th
element
from
the
end
|
zero
its
density varies from
intermediate between
(
*
"
to
(7+la\m
"
*"
n
)
)
Its
.
is therefore and
coa**1-
**
10
Thus the
INTEGRAL
CALCULUS.
mass
of the whole
rod lies between
and and in the when limit, increases becomes indefinitely,
n
ra+1
5. Determination
of
a
Volume
of Revolution.
formed to find the volume by required axis the revolution of a given curve about an AB in its own planewhich it does not cut. the Taking the axis of revolution as the cc-axis, The in Art. 2. be described exactly as figure may
Let
it be
Fig. 3. trace in elementaryrectangles AQV P-fy^P2Qz"etc., and their revolution circular discs of equal thickness, of volumes L2 LQ19 "jrA nrP^ Q", etc. The several annular portionsformed by the revolution of the portionsAR^^ P^R^P^ P2E3P3, etc.,may be con. .
12
Then have
INTEGRAL
CALCULUS.
dividingas
before
into
re
elementarycircular laminae, we [Art.4.]
y^dx /c
"
4a:r
/ xdx
2
AN PA7' and
of =J cylinder
as [Or if expressed a
radius
heightAN.
series dx
Volume
=
4a?r
I
o
[c
x
r
=
"
.
2a?rc2.]
of the
=
[Art.4.]
formed spheroid prolate
l about
2
Ex.
2.
Find
the volume
by
the
revolution
of the
~+^ellipse
2 *
the #-axis.
Fig. 5.
Dividing as
axes
before the
coincide with
elementary the volume #-axis,
into
circular laminae is twice
whose
/ Try^dx.
Now -a2
a
-
x*)dx
to Article 4, is equal to which, according
5[a*.("-0)-^]
or
and
the whole
volume
is
NOTATION,
SUMMATION,
obtain the
APPLICATIONS.
13
[or if desirable we may the sign of integration, as
same
result
without
using
EXAMPLES.
1. Find
the
area
bounded #=".
by
the
curve
y
"
^^ the #-axis,and
the ,#-axis find the
the
ordinates
area
#=a,
2. If the
in
Question
method
1 revolve
round
volume
of the
solid formed. the of Art.
3. Find
by
the
2, the
area
of the when
the
x
=
triangle
a.
formed
by
Find
line y=x tan 0, the #-axis and of the cone formed also the volume the #-axis. of the
line
this
triangle
revolves
4. Find
about the about the the
volume the
revolution
cut
y-axisof
of the
by the reel-shapedsolid formed that part of the parabolay^"^ax by
the revolution of
off
by
latus-rectum. volume
=
5. Find
sphere formed
.r-axis.
the
circle
x2+y2
the
curves,
a2 about of the the
the the
6. Find
areas
figures bounded
the ordinate
area
by
x
=
each
of
the
following
volume
#-axis, and
revolution
=
formed
by
of each
about
h ; also the the #-axis :
(a)
7/3 a*a
(8)
7. Find each the
mass as
aty
a
of the of
circular
disc from
of the
which
centre.
the
density at
by the #-axis,supposing
point
varies the
distance the
8. Find
mass
prolate spheroid
=
formed the
revolution the
of the
l ellipse^2/a2-f^/2/62
about
density at
each
point to
be //x,
CHAPTEE
II
GENEEAL
METHOD.
STANDAED
FOEMS.
6.
the
Before
proceeding
Calculus,
will the in
we
further shall
cases
with
applications
a us
of
Integral
which of result
establish enable
general
to
theorem the
many
infer
operation
n
indicated
by
I
a
"p(x)dx
without often
recourse having difficult,process
to
the
usually
or
tedious,
and
of
Algebraic
Trigonometrical
Summation.
7. finite and PROP. and b of
Let
"/)(x) be
any
function
of
x
which values
is
a
continuous the b
"
between
x
given
a
finite
and
variable
a
;
let
be
"
6,
n
suppose
the
each
difference
to
be
a
=
divided nh. series It
into is
portions
to
equal
limit
h,
so
that
sum
b
"
required
find
the
of
ike
+
of +
the
ft[0O)
when increased
"p(a + h)
diminished
4"(a + 2h)+...
+
0(6
and
-
h)
+
0(6)],
n
h
is
indefinitely,
therefore
without
at
once
limit.
be
seen
[It
be the
may
that
sum
this
is
limit
is
finite, for
if
"$"(a+rh)
greatest
term
the
-
+ rh) a)"t"(a
+
+ h"$"(a
GENERAL
METHOD.
STANDARD
FORMS.
15
of
which
x
is
is finite four all values since by hypothesis""(#) finite, between b and
intermediate
a.]
of
x
Let
be \fs(x)
another
function
such
that
is "j)(x)
i.e.such its differential coefficient,
that
We
shall then
prove
that
Lth^["fa)+^a+h)+^
By
and
where
definition
^a)*
a
=
therefore is
a:
diminishes
quantity whose indefinitely ; thus
a
"
limit
is
zero
when
h
h(j)(a) Similarly h"f"(a
=\/s(a+ 7i) t/r(a)
+halt
etc.,
"
+ nil) Ih) \[s(a
=
"
\
where
the
quantities a2,
limits
a3,
...,
an
are
all, like
av
whose quantities
are
zero
when
h diminishes
indefinitely. By addition,
+ 0(a + h)+ "f"(a h["f"(a)
Let then
a
be the
of greatest is
the
quantities av
i.e.
a2,
.
.
.
,
an,
Afoi+ag+^.+On]
and therefore vanishes
"nha,
"("
"
a)a,
in the limit.
Thus
16
INTEGRAL
CALCULUS.
limit zero; hence if desire,it may be added to the left-hand member and it may then be stated that this result, The
term
is in fc"/"(6)
the
we
of
.e.
1 "/)(x)dx \ls(b) \ls(a).
=
"
This
result
denoted is frequently \[s(b) \fs(a)
" .
by
the
notation From the
p\/r(a3) J
that when is "/"(x) process obtain
rb
this result it appears function ^fs(x) (of which
the form
of
the of
differential
is obtained, the coefficient) summation trigonometric avoided. The letters b and
a are
or algebraic
to
I
a
"j)(x)dx may
be
to denote
supposed in finite quantities.We shall
as
the above work extend our now the limit when
notation
so
to let
I "f"(x)dx express
a
"
b becomes
i.e. of ^(6) ^js(a), infinitely large I
a
Ltb=x I "j)(x)dx. (j)(x)dx
=
a
(j)(x)dx fb
we
shall be
understood fb
to
-\I,(a)]
or
Lta=00\ "f"(x)dx.
Ex. Hence
1.
The
differential have
coefficient of ^"
-
is
plainlyxm.
if
we "$"(x)=xm
and "df(x)":L^ ' m+l
J
/
x
m+1
m
+ \
m+l
GENERAL
METHOD.
STANDARD
FORMS.
17
cos a?
Ex. 2. The quantitywhose Hence known to be sin x.
"6
differential coefficient is
is
cos
x
dx
=
sin b
"
sin
a.
Ex. 3. itself ex.
The Hence
quantitywhose
differential coefficient
is e* is
Ex. 4.
EXAMPLES.
Write
1.
X
down
the values of
2. I X 2i,
0
/V"dr, CiX) /b
a
it
/V"cfo?, 3.
Cf/JCm
o.
ir
rl
,-2
I
X
cfo, d/X^
4.
1
/2
cos
rA
r4
x
dx,
6.
/
sec2^;dx^
7.
/
o
\
ia
8. Geometrical
The
Illustration of Proof.
above theorem may be
proof of
thus be a
:
"
the
interpreted metricall geo-
of a curve of which the ordinate is finite Let AB portion and continuous all points between A and B, as also the at makes tangent of the angle which the tangent to the curve with the a?-axis. Let the abscissae of A and B be a and b respectively. Draw ordinates A N, BM. be divided Let the portionNM into n equal portionseach of lengthh. Erect ordinates at each of these pointsof division
cutting the
curve
in
P, Q, R,
...,
etc.
tangents AP^
to parallel
and PQi, QRi, etc.,
the
Draw lines
the
successive
AP2,PQ2JQR2,...,
the let and ,r-axis,
=
let the
equation of
the
curve
be
y then
=
and ^r(x\
V^') ""M" + Zh\ etc., are + h\ "$"(a "f"(a\ "$"(a respectively
tanP.JPj, taii^Pft,
B
etc.,
E.
I. C.
18
and
INTEGRAL
CALCULUS.
the lengths respectively
-h),
...,
are
Now
it is clear that the
sum algebraic
of i.e.
P2P, "2", R2R,
Hence
is
...,
MB-NA,
s,
u
L
M
x
Fig. 6.
Now
portion within square brackets may be shewn for instance be For if R^ with h. diminish indefinitely the sum of the several quantities PjP, Q^ etc., greatest
the
to
the
[P1P+Q1Q+...]
But if the abscissa of
is
"nR1R,
i.e.
"(b-a)-}~.
Q
be called #, then
and
+ (x)
+ -^"(x
Qh\
[Diff.Calc. for Beginners, Art. 185.]
so
that
R^R
=
4- "9A) "(x
=
(x+ Oh),
and which is
(6 a)
-
an
infinitesimal in
of general
the first order.
20 10. When
we are
INTEGRAL
CALCULUS.
is not specified and the form of the (at present) merely enquiring
a
the
lower
limit
function \fr(x), unknown differential coefficient whose is the known function $(#), the notation used is
the limits
beingomitted.
11. Nomenclature.
The nomenclature
b
of these
or
is as expressions
follows
:
r(p(x)dx
is called the "definite" b ; a and
of "f"(x) between integral
limits
fx
I where
a
or "j)(x)dx
\[s(a) \{s(x)
"
the upper limit is left undetermined "corrected" integral;
is called
"f"(x)dxor
without
as
-^(x)
the
limits and regarded merely specified any of reversal of an operation the differential is called
an
calculus
"indefinite"
"
or
unconnected
"
integral.
12. Addition It will be
of
a
Constant.
that if is "p(x)
the differential coefficient of ^]s(x\ it is also the differential coefficient whatever of \lr(x) C is any constant + C where ; for is zero. the differential coefficient of any constant obvious
we Accordingly
might
write
This
constant
is however
not
written clown, usually
GENERAL
METHOD.
STANDARD
FORMS.
21
but
will be
understood
to
exist
in
definit all "cases of in-
integration though not expressed.
13. Different
will
of indefinite processes give results of different frequently
integration
form
;
for
instance is the
Idx }*/I-x2
,
is sin'1^
or
"
cos~%, for
Vl
pressions. ex-
,
differential coefficient of either of these Yet it is not to be inferred that sin"1^^
"
cos'1^. sin"1^ and
"
But
what
a
is
reallytrue
is that
cos"1^
differ by
constant, for
so
that
f
"7
1
dx
=
Vl-a2
,
sin ~lx
or
dx=
"
cos-^
J/s/l-a;2
the
constants beingdifferent. arbitrary
14. Inverse
Notation.
verse notation for the inAgreeably with the accepted tions, funcand inverse Hyperbolic Trigonometrical we might express the equation
or
j5^)
it is useful occasionally
=
^);
to
and
employ
this
notation,
22
INTEGRAL
CALCULUS.
character which very well expresses the interrogative of the operation are we conducting.
15. General
Laws
satisfied
by
the
Integrating
Symbol \dx.
(1) It will symbols that
be
plain from
the
meaning
of
the
s
but that
constant. is "j"(x) + any arbitrary (fi(x)dx l-y-
(2) The
for if u, v,
is distributive; operationof integration
w
be any
functions of
x,
-T-j |u^+l^^+l^^r
and therefore
constants) (omitting
=
\wdx l |i;cfe-f JurZ#+
of integration is commutative (3) The operation with regardto constants.
(I'll
For
if
-j"
=
v,
and d
,
a
be any du
constant,we
have
so
that
(omitting any
constant
of
integration)
au
=
\avdx9
or
a\vdx=\avdx,
establishes the theorem.
which
GENERAL
METHOD.
STANDARD
FORMS.
23 of
16. We several
proceed to a detailed consideration forms of functions. elementaryspecial
now
17.
Integration of
xn.
By differentiation of
d
_
-
"
n
+ 1
nrfll
^
we
obtain
xn+l
_
dx
n
+ l
seen
(as has Art. 7, Ex. 1)
Hence
been
already
in
Art.
4
and
in
Thus
power
the of
x
rule
for the
Increase is,
so
of any constant integration the index by unity and divide
by
the index
increased.
For
example,
/nA
X X
r
=
5
Q
11
r
x~^
1
~"''}X x=if
EXAMPLES.
;
}x
T4=
"4^4-
TTn'^e down
the
of integrals
1.
x
1
0
"#"
^7999 #1000.
O
^"J
""}
-^"Jf
b
c
a
O.
24
INTEGRAL
CALCULUS.
18. The
Case
of x~\
that x~l Thus
or
-
It will be remembered ential coefficientof
is the
differ-
x
logx.
fl
Jx
\-dx
This therefore forms
f
an
=
logx.
to the apparent exception
^n+l
rule general
\xndx
=
19. The
Supplyingthe
result, however, may be arbitraryconstant, we
deduced have
as
a
case. limiting
/xndx
=
+
C
=
~
"
-I-A
n+l
n+l
where
A
=
C+
"".
n
+ l
and
is still an
constant. arbitrary
Taking the
limit when
n
+
l=0,
the form
-
-
takes
logx,
Calc. for Beginners, Art. 15.] [Diff.
and
as
C is
we arbitrary
may
-
suppose
that
it contains another
a
tively nega-
infinite A. portion
portion
-
togetherwith
arbitrary
7i
~\~J.
Thus
Ltn==-i {xndx logx + A.
=
20. In
we
the
same
way
as
in the
of integration
xn
have
1
=
+ b)n (n + V)a(ax
and
+ 6) ^-log(a% "v
=
"
GENERAL
METHOD.
STANDARD
FORMS.
25
and
therefore
f/
y
\(ax+o)ndx
IA
7
+ ")n+1 (oo5
'
=
"
"/
(n+l)a
,
and
I '"r Jax + b
we
=
-\og(ax + b\ 6V
a
f fFor convenience
shall often find
Jax
1
+ b
"jdx
printed as
Jax
"
I
+ b'
-r,
Jja2+x*
EXAMPLES.
,
dx
as
I J*Ja* +
,"
x*
o,
etc.]
Write
down
1. ax,
a
the
of integrals
a
"
of1, a+x,
x_ a+x a? x
'
x,
a"x?
1 a-\-x
2
x
3.
a
+ x
a
"
bx
(a-#)2" (a x)n*
"
a+x
a-x*
x+a
x
"
d
(a+x)2
(a
"
xY
21. We
may
next
remark
that since the differential of
coefficients of
and ["f"(x)]n+l
log$(x)are respectively
and
we
have
{["t"(x)]n"t"'(x)dx
=
and
is of great especially the integral It may thus : be put into words use. of is the differential of which the numerator any fraction is coefficient of the denominator log (denominator). second results
"
The
of these
For
example,
INTEGRAL
CALCULUS.
/ /co\,xdx J
=
*"
xdx
X
=
.
log sin log cos
#,
Sill
/tan
.#
dx
=
"
I
"
"
-a?^
=
"
x
=
log sec
x-
J
cosx
EXAMPLES.
Write down the of integrals
nex,~-,
" Ct ~\~
(a
22. It will the that
now
be
that perceived
the
of operations
Calculus are of a tentative nature, and Integral in integration ledge success depends upon a knowof the results of differentiating the simple
It is therefore forms
same
functions. of standard
which list the
necessary is now
as
to learn the table
the practically
and differentiation,
that proofsof these
appended. It is already learnt for
results lie in several and
a
of the the righthand members differentiating results. The list will be gradually extended list givenlater. supplementary PRELIMINARY TABLE
TO OF
RESULTS MEMORY.
TO
BE
COMMITTED
28
INTEGRAL
CALCULUS.
them.
as x
The
reason
is obvious. the
Each
of these
functions
decreases
efficients co-
increases
are
through
therefore further and
a
first
quadrant
; their
differential of each
" "
Also For
it is
a
negative. help to observe
the dimensions
" "
side.
zero
instance,x
being supposed linear, /
"
is of
to
i
v
a
"
X"1
dimensions.
There
C
could dx
therefore
be
no
-
prefixed
a
the
in-
tegral. Again
/
J
2 d~"
-\-X
2
is of dimensions -1.
-1.
Hence the The
the result of
integrationmust
not
be
of dimensions is of
zero
Thus
integralcould
student should the
be
tan"1a
(which
no
dimensions).
therefore factor
-
have is to be
in remembering difficulty
in which
cases
prefixed.
a
EXAMPLES.
Write
:
"
down
the
indefinite
integrals of
the
tions following func-
'
2.
3.
*
cos2-,
2 cot
x
coss#
.
sin #,
4.
+ tan
x,
cos^f
.
-+-^snr^/
V
\sin^7
#e + e*
'
log sin
x
^sec-1^.
\/^-i*
CHAPTER
III.
METHOD
OF
SUBSTITUTION.
25. The
to
z
Change
of
the
Independent
may
Variable.
be
independent
by
the
variable
x
=
changed
from
x
change
F(z), by
the
formula
V
being
Or if
any
we
function write will be
of
x.
V=f(x),
the
formula
To
prove
this, it is only
necessary
to
write
u
=
\Vdx\
then
=F.
du But
--
du
= "-_-=
dx
dz
~rrdx
V-j-9
dz
dz
dx
whence
u
I
=
J
-j-.
dz
30
INTEGRAL
CALCULUS.
Thus
to
/ integrate
1
-dx, let tan~1#=;s.
dx
n
Then
and the
becomes integral
*"
dz
26. In
usingthe
formula
after
the choosing
to
form
use
it is usual
make
of the transformation x F(z\ of differentials, writingthe
=
equation
j^=F'(z)dx
as
=
F'(z)dz]
the formula
will then side
of the left hand
Thus may in the
dx reproducedby replacing and x by F(z). by F'(z)dz,
be
we precedingexample,after puttingtan~1^ti=0,
write
*=d*
-
and
l+.r
I+x*
27. We
is
a
next
one
consider the between
case
when
the
integration
definite
limits. specified
x
=
The
result obtained
above, when
F(z) is
Let then and if the limits for
a;
be
a
and
b,we
have
METHOD
OF
SUBSTITUTION.
31
Now and Also
when
x x
=
a,
z z
=
F~
when
=
b,
=
F~
\a) ; \b}.
f{F(z)}=-j^,{F(z)}
and
whence
so
with rethat the result of integrating gard f{F(z)}F'(z) limits F~\a) and F~\b) is identical to z between with that of integrating f(x)with regardto x between the limits
Ex.1. Let
a
and
6.
Evaluate
/
-
cos
\Txdx.
;
J
N/a?
dx=Zzdz 2-2(^2=2
x"z^^ and
-
therefore I -cos2. J z
cosfjxdx"
^x
/cos
J
z
dz
=
2
smz
Ex. Let
2.
Evaluate
/.Aos
x^dx. 3xPdx=dz
and ^3=2;,
/.
therefore
;
/^2cos x*dx
=
llcoszdz ^smz=^
=
sin x\
Put
^?=tan
e, then dx=sec20d6
when when
#
=
;
0,
1,
we
have have
0
=
0,
4
#
=
we
0=.T ;
32
INTEGRAL
CALCULUS.
ir
ir
:.
f-T^=dx \J\+*
=
P *?"| sec20 dB
=
{
=
sec#
{
-
B tan fsec
0 dB
fsec 0 "
dx
x _x
=
sec
-
sec
0
=
V2
-
1.
Ex.4.
Evaluate
f
"4
exdx
=
\ Tsech^^]. [i.e.
e'
jo
When
Let
3=e.
6^
=
^, then
dz.
#=0, 0=1, and
when
#
Hence
=
rtan-'/V= tenL
""
-
tan-1
=
t
2
"
Ji
o
The
indefinite
is tan~V. integral
EXAMPLES.
1.
Integrate
excosex
(Put
^=4
-
cos(log x)
(Pat logx
I*
=
4
"
x
2. Evaluate
J 1+^4
acos#
\-=-,dx (Put x*=z\
"
-
J l-f#6
(Put a*=z). v
x.
reintegrate+ fl ^ Evaluate
"
-,
a^sin
ex + b tanh
a
(Put ^+1=4
Q
5. Evaluate
/"
dx
-
(Put (Put x-\=z). (Put ^=02).
6. Evaluate
/
a*
7. Evaluate
/*"
"da?
-1
J 2V^(1+^)
8. Evaluate
[
/
J 2W#
dx.
9. Evaluate
dx.
-
1
METHOD
OF
SUBSTITUTION.
33
NOTE
ON
THE
HYPERBOLIC
FUNCTIONS.
28. Definitions.
it is desirable that the purposes of integration the definitions and student shall be familiar with of the direct and inverse fundamental properties functions. hyperbolic values of the By analogy with the exponential functions the exponential cosine, etc., sine, tangent,
For
"
_
"
e-*
~
ex+e~x
_
__
ex-e~x
PTP
2
are
2
e?+e-*'
written respectively
cosh x,
tanh#,
etc.
29. We
Elementary Properties.
have clearly
tanh
-,
C/
~
V
X "C"JLJ.iJ.l
x
=
-
coth
x
=
"
e~x
sinho?
t"rihx
-=cosh
^
2 sinh
x
cosh
x
=
2
"
:
^
"
"
-^"
.
=
^
"
=
sinh2#,
common
AM
with many other results analogous to the formulae of Trigonometry.
E.
i. c.
c
34 30. Inverse Let
us
INTEGRAL
CALCULUS.
Forms. for
the
search sinh"1^.
meaning
of
the
inverse
function Put
t
then
x
=
smh
y
=
and Thus and
we
"
=
x"
y
=
log(x"
a
shall take this
with expression
sign, positive
viz., + #2) as sinh"1^. log (" + "v/l
31. cosh~1x Similarly, putting
x
"
=
y,
we
have
ty+e-y cosh y
==
"
JL
and and whence and
we
e*y
ey
=
x"*Jx*-l,
log(x " *Jxl 1),
"
y
=
shall take this
with expression
1
a
positive sign,
viz.,
/#2
"
as
32.
Again,puttingi"nh-lx
x
=
"
y,
we
have
tanh
y
=
-
and
therefore
e2y=-"
1"
x
whence
tanh
-
lx
=
4-log S ^
"
"-
1"
05
36
INTEGRAL
CALCULUS.
x
cu-l
=
whence
tan-
0, 2
x
eu-l
^TTT
=
tan2
2"
and
tan
a;
=
.e-
0_e-
!~4^-: ~2~
Hence
^
=
tan
~
1sinh u
=
gd u.
Thus
logtan(j+|)=gd-^
of
x.
the inverse Gudermannian
EXAMPLES.
Establish the
1.
results : following
"
/cosh#cfo?=sinh#.
4.
/cosech2.rc^= -coth#.
2.
/sinh
J
xdx
=
cosh
x.
5.
J cosh%
/sm,.^o?.r
=
"
sech
x.
3.
6. (sech2A'dr=taiihtf.
J
7. Writing sg results :
"
x
for sin
gd x, etc.,establish
the
following
(a) /
METHOD
OF
SUBSTITUTION.
,*
37
36.
Integralsof
and
The
differential coefficient of
X
-4-f
-
loge"
is
* =
log"
a
,
"
=
smh
.
,
-
1 l.
z
a
F
dx
=
..
bimilarly
37. In resemble and the
0.
.,
,
x
--
+
\/x2
"
a?
Jx/^2-^2
the inverse for the
log
^ =cosh-1-.
,
a
a
forms hyperbolic
these
^,
results
that
integral I
Va2-*2 analogyis an might have
dx
t
=.
,
sin'1-* viz.,
a
aid to the memory. established the results thus
:
"
38. We
f To find I dx Hence
)*Jx*+a^
=
put
oj
=
a
sinh it, then
a
cosh
u
du
and
\/x2 +
=
az
=
a
cosh
u.
=
W^2
+ tt2
J
leZu t6
= =
sinh"1-.
a
a? Similarly a cosh putting Fa sinh u f dx I!" ri Jx/x2" a2 J " sinh
=
" -
u,
we
have
du
=
f
J
/
7 laK
=u
=
cosh~1-.
a
11^
a
u
Integralsof
39. To Let then
integrate *A2-tf2.
a?
=
a a
sin 0 ;
cos
dx
=
0 d$,
38
INTEGRAL
CALCULUS.
and
{+/tf-^dx
=
ia sin 6
.
a
cos
0+
-^
or
-
sina
40. To Let then
,
integrate
cc
=
a
sinh z,
cf^ since
1 + sinh20
=
acosh0
then
=
cosh2z,
we
have
I J^^dx
=
a2\ cosh2z
dz
=
|a sinh
0
.
a
cosh "z+ -^-
.....
.
Va2
.
,
2-smh-1x
METHOD
OF
SUBSTITUTION.
39
41. To Let then
then
integrate
x
"
a a
cosh z, sinh
z
dx since
cosh%
"
=
dz ;
1
=
sinh20,
a2 sinh20 dz
Jsif^cPdx J
=
j
C62?
=
Ja
sinh
z
.
a
cosh
0
"
^-,
\2
or
"
a
a2!
-log-
we
42. If have
we
put
tan#
=
", and
therefore
__
[by Art. 40.]
tan
"
x ^
sec
05
-
,
h
,
a? + J log(tan
.,
sec
^),
snce or
_ _
+
2 cos2#
J log^
n
-,
"1"
40
INTEGRAL
CALCULUS.
43.
Integralsof
^ 2t
=
cosec
x
and
sec
x.
Let tan
z
;
differential takingthe logarithmic
9x
^
1
-
dz
7
=
"
dx
or
-;
"
dz
"
x
2
^dx
.
z
2tan2
Thus
n,
smx
z
I cosec
xdx"\
"
=
logz
=
logtan
^.
In this Then and
example let x
dx
=
=
-=
2*
+ y.
dy, logtan logtan
+ 9)(T
sec
ydy
=
Hence
Isec xdx
=
r+s) (
or
""
~
*x-
44. We
have
now
the STANDARD
ADDITIONAL f dx
. ,
FORMS,
x+\/x*+a?
g
a
Jx/^+o1
f
dx
,x
JTP^l
\\/a2 x2dx
"
=1"g
=
I+/x2 4 a2 die
=
l+Jx2 a2dx
"
fi2
=
2G"
a
METHOD
OF
SUBSTITUTION.
Icosec
x
dx
=
logtan^.
Isecsccfo =log tanf^-
EXAMPLES.
Write
down
the
of integrals
3.
x
x
4.
5.
7.
1 8.
cosec
2#, cosec(a#+"),
-sin2^'
1"
tanV
3sin^-
^'
10. Deduce
"/
a
sin ^7+ 6
cos
x
/cosec#cfo?=logtanby expressing cosec
-
x
as
^j
11.
Find
\SQexdx by puttingsin
x=z.
42
INTEGRAL
CALCULUS.
12.
Show
that
/ /
sec x
dx
=
cosh
13.
Integrate
1
tflogtf'
when
lrx
represents
log
log
log
...
the
^7,
log
being
repeated
times.
r
15.
Prove
[ST.
PETER'S
COLL.,
etc.,
1882.]
44
INTEGRAL
CALCULUS.
with
a
new
\ dx integral 0'(#) lx/r(#)cfo
be
more
which
may
than integrable easily
the
original
product.
be put into words thus rule may of the product "j"(x)\{s(x) Integral 46. The
=
:
"
1st function
-the
Ex. Here another
1.
of 2nd Integral of [Diff. Co. of Integral
x
cos nx.
1st x Int. of
2nd].
Integratex
it is
important
which
to
connect
if
x
possiblejxcosnxdx
has been function in removed. Then
with This
in integral
the factor
if x be chosen the be done as may second "$(x\ i.e.unity,occurs integral
$(x\ placeof x.
since in the
Thus
by
the rule
(xvxnxd**,*!*"?J
'
/"l.5
J
n
sin
"
9^7
If
"~~~~\
cosn"N
-
x
-
I
n /
n
n\
cos nx
sin
'
nx
47.
an
Unity may integration.
be taken
as
one
of the factors to aid
Thus
/logxdx"
=
/1 logx
.
dx
x
logx
log x
"
/x
"
x)dx -(log
=x
"
I \dx
INTEGRA
TION
B T
PARTS.
45
48. The
operationof integrating by parts may
times.
? dx
be
several repeated
mt.
Thus
J
f 9 / #2cos
#2sin
"
-
nx
"
nx
f sin / 2#
0
-
nx
7 dx.
n
J
n
and
n
/
finally,
Hence
f
J
$x^ J
nx
dx
=**** "
n
-*\_^COS
nL
n
#2sin
_ ~
nx
Zx
I
cos T9
nx
2 sin
7^
nx
*
of the subsidiary into 49. If one returns integrals form this fact may be utilized to infer the the original result of the
Ex.
1.
integration.
dx
=
/eaxsin bx
/eaxcos
and
"sin
bx-~\
e^cos
bx
dx,
and
bx dx
=
"cos
bx+-l
e^sin bx dx ;
if P therefore,
Q
stand
for respectively and
/eax$m
we
bx dx
/eaxcos
bx,
bx
dx,
have
aP
+bQ
=
eaxsin
and
whence
-bP+aQ=eftxco8bx"
n P=eax-
nrct
sin bx
=
"
"
b
r?
cos
bx
/
a2+o2
and
w+v
"
(a2 +
(bx 62)~Yeaxcos
\
"
tan"1
-
).
aJ
46
The
we
INTEGRAL
CALCUL
US.
student
will observe
that
these
"
results
are
the
same
that
should
obtain
by puttingn=
I in the formulae
^)""ss""it^"^(^+"^)'
Gale, for Beginners, Art. 61, Ex. 4.] [Diff.
And
pnsi^
this
/
is
otherwise
game as
obvious.
For
a
""
if to
differentiate
62 and
to
rxsm/j^,\ jg fag
increase the
to
multiply by
r
""
factor
Va2 +
angle by tan"1-, the
a
which integration,
is the
inverse
and
must operation,
divide
out
again
the
factor
Va2+62
diminish
the
angle by
xl
tan"1-.
Ex.
2. r
\/a2 Integrate
"
by
the rule of
by parts. integration
J A/o2^2^=
,
_
_
[Note this step.]
%
c
-
a2sin~
l-i
Iv a2
J
"
CL
and dividing whence, transposing by 2,
which
agrees
3
with
the result of Art. 39.
Ex. Here
e*xsm2x Integrate
e3xsin% cos3#
=
cos3^.
"
"
"(1
"
cos
x 4#)cos
-=
-x _(2e3a:cos
-
INTEGRA
TION
B Y
PAR
TS.
47
Hence, by Ex.
I e^siiA
1,
cos3# dx
"
^
"\
-j=
cos
fx
"
tan~
1r j
J_
-
3\/2 ,: V
cos(3^-^--^cosf5^-tan-1|) V 3/J 4/
^34
[Compare
n=
"
Ex.
16, p. 55, Diff.Gale, for Beginners, putting
l in the
result.]
EXAMPLES.
Integrateby parts : 1. xex, x^e*,xze?) x cosh #, ^?COS2 2. ^?COS^7, ^2COS07,
3. 4. 5. 6.
x
sin
x
cos
#,
^
sin
x
sin 2# sin 3^.
#2logtf, ^n(log^)2. ^nlog^7, e*sin x cos ^ cos 2#. e^sin^costf, eaxsin^ sin qx sin r^?.
7. Calculate
|^sin^^,
0
/*x sin2^pc?^1, /
0 0
8. Show
that
9.
Integrate Isin"1^^,/^sin"1^^, \
50. Let
axes
Geometrical
Illustration.
and
referred to be any arc of a curve Ox, Oy, and let the coordinates of P of Q (xv y^).
PQ
rectangular be (XQ, yQ),
the
Let
PN,
QM
abscissae of the
area
PNV QM1 pointsP, Q. Then plainly
=
be
the
ordinates
and
PNMQ
rect.
OQ
-
rect. OP
-
area
But
area
PNMQ
=
f
48
INTEGRAL
CALCULUS.
and
area
PN^M^Q
=
I x dy.
cv\ o o
Thus
ri
J
Let
us
now
consider the
curve
to be defined
by
the
equations
and and let y
t0and t"be
the values We then have
the values of t corresponding to #0, y0, and a^, 2/1 of cc and y respectively.
ri
ri
7
I
*0
2/"x^=l vdu=\
*"
"0
,
r*i
and
I o?c?2/=l udv**\
and
INTEGRA
TION
B Y
PARTS.
'*
49
so
that the
above equation
may
be written
and thus the rule of
by partsis established integration
geometrically.
51.
Integralsof
the
nx
Form
nx
I a^sin
Reduction
#mcos dx, I
dx.
for such integrals the above as be found. Denote them respectively readily by may have at Sm and Cm. Then, integrating we by parts,
once cos nx m~
formulae
and(7m=
Thus
and
Om=
cosnx
=
,sin7i#
,
"_ ---
____+m^
771(971 v___
"
and
Cm=
n
"
-l"-
,cosnx
----
m(m
"
1)
"
Thus and m
when
=
the four
for integrals
the
cases
m
=
l
are
found, viz.,
"Sn
"
n
I sin
nxdx=
D
^
cos n
me
,
J
E. I. C.
50
INTEGRAL
CALCULUS.
GT0
=
xv
f
I
cos nx
sinrac
dx
7
=
,
at. o\
=
~
eosnx
Icesin
nx
dx
7
=
"
x
"
sinnx \"
,
t
0
=
\ cos \x J
can
^
T^CC ax
=
all others the above
be deduced
by
successive
of applications
formulae. of the
52. Extension
Rule
of
for
Integration by
dashes denote
Parts.
If
to
u
and
v
be
functions suffixes
x
and
differentiations and
x we
rule for
with respect integrations of the extension prove the following may by parts, integration
=
\uvdx
uvl
"
u'v
where
u^n~1^ is written
for =uvl
u
with
TI
"
1
dashes; for
\uvdx
"
\u\dx,
Vufv^dx =u'v2 "\urfv^dx, \vtf'v2dx =u"vz
=
"
Vuf'Vzdx,
"
I u'"vBdx ufv^
etc.
=
I u^'v^dx,
etc.
Iu(n l)Vn _1dx
-
=
u(n~ Vvn
-
I vf^Vndx.
52
INTEGRAL
CALCULUS.
we
have bx dx
=
"
Ixneaxsm
eaxsin (bx
r
"
J
d")
"
^ r2
eaxsin(bx
"
r3
^^~
n\
or
eax{P sin bx
"
Q
cos
where
X
-
3-
COS
30"
"
...
xn
xn~l sin 0
"
xn~^ sin
Q=
"
n
"
^-
20
+
n(n
"
1)" ^- sin 30
...
Similarly
L^a*cos ix
Ex.
1.
dx
=
eP*{Pcos bx+Q
sin
bx}.
Integrate ix^smxdx.
\e*smxdx
"
Since
S^e^sinf .r
-^J,
-
we
have
f^3ea:sm^^=^32'^ea;sm('.r ^ 3^22~Vsin^
-
2"
.
VVsinf ?" 4
.77
-
6
.
2~VsinCr- TT)
\
/
=etc.
Ex.
2.
Prove
^|
^Vto^ito-^-iy^j^^ /r=n
EXAMPLES.
1.
-rjQ ^s
Integrate (a) femai"~lxdx.
(d)
/"
(5) (sfaitr^xdx.
(c)
(e) \
Ixv"Pxdx. '(/) /"cos-1^.
INTEGRA
TION
B Y
PARTS.
53
2.
Integrate (a) [x
dx. sm"1f
(c) /sin-1'
(")
/^5^"'.
tan
-
(d)
/ptn
~
lx
r
pin tan
-
lx
-dx.
(c)
J
*-*
dx.
dx.
4.
Integrate (a)
../*..
I
...
.
r
..
e(suix + cosx)ax.
(a)
\x
(b) I xefsm^x (c) I cosh
5.
dx.
(e) I^22*sin (/) /cos
2." dx.
ax
sin bx dx,
b log"\dx. -j
Integrate
/log
-
sin'1^ dx.
J
x
6.
Integrate
7.
Integrate
8.
Integrate
(d) Integrate
/cos 201og(l+tan 0)dO.
9.
J*4"|^.
1-cos^ /" c?2v
"
^^
TKJPOS) 1892"]
(")
i
[a, 1892.]
c?v
"
"
10. Prove
/\
T-"
that
j
i
/u
2dx=u
7
du
v
"
+
/ v-"dx.
C d^u
i
11.
Integrate
/(asin%
+ 26 sin
x
cos
x
+
c
cos2^)e*^
[a,1883.]
54
Show that
INTEGRAL
CALCULUS.
12.
if
u
be
a
rational
function integral
of x,
where
the
series within
the
brackets
is
finite. necessarily
[TRIN. COLL., 1881.]
13.
If
u"
Ieaxcos bxdx,
v
"
Ieaxsm
bx
dx,
prove
that
and
that Prove that
+ "2)0*2 + v2) (a2
=
era*.
14.
-"
m+1 Also that
m+L
(m+1)2
3
^"-1 where
15.
(-ir-^!?
I stands Prove
for
logx.
that
(i.)
{e^w
J
+^""j)"2 leax^n-'2bxdx. J
a?+ri2b'2
[BERTEAND.]
16. Evaluate
/x* log(l x^dx,
-
and
deduce
that
iT5
+
277
+
3T9
+
-==9""310ge2'[a,1889.]
CHAPTER
V.
RATIONAL
ALGEBRAIC PARTIAL
FRACTIONAL
FORMS,
FRACTIONS.
ALGEBRAIC
FRACTIONAL
FORMS.
54.
Integration
of
-
-"
or
"
a?
^
\
forms
and
-"
-
-9(x"a\
into
Partial
*"2
Either Fractions.
of
these Thus
should
be
thrown
=___
_
x2
a2
"
2aj\x
"
a
x
+
a
1,
=
a^
"
a
;"
-
F
=
"
1
.1 coth"1
i^"l
a
s-
log
"
"
2a
#
4-
a
L
a
J
f
^ a;2
=!f(-J-+_J_Yfo
2aJ\a4-a)
a
"
Ja2"
x/
1
,
a+oj
F
a;
l,i
=-tanh"1-
T^l
.
=
^"
losr"
toa
"
2a
La
aj
56
INTEGRAL
CALCULUS,
Compare the (
with
.
forms
of the results in square
1
brackets
the result before tabulated C dx
"
for
x\
-=
viz.,
1
= o
-
Jcr+or
!~n
tan'1"
a a/
)
dx 55.
Integrationof
Let
f-H
a
J
f
6
c
2
=1f.
a
dx
J
\
a^a
AV_^2a/ 2J
"
4a2
dx
or
2
we
take
as
the former b2 is " or
"
or
the 4ac.
latter
arrangement
cording ac-
"
Thus
if 62
4ac,
or
coth"1"
7_
.
;
If b2
"
4"ae,
I
=
"
/
tan
"
l"
-.-
or
--
"
cot
~
These any
but differ at most by constants, expressions givencase a real form should be chosen.
in
RATIONAL
ALGEBRAIC
FRACTIONAL
FORMS.
57
56.
of of expressions Integrals
px + q
the form
*
can
be obtained
at
once
by
px + q
~~
_p
tion followingtransformapb 2a (2ax+b) the
,
the
of integral
the first part being
+bx+ ^" log(ax2 Za
and
that
c),
of the second
part beingobtained
notice how
by
the last form is
article.
[The beginnershould
obtained. shall Jirst fraction
the above
It is essential that the numerator be the that all the #'s of the
of the coefficient of the differential
numerator
denominator, and
are
therebyexhausted.]
T?
'
=
+ J log(^2
4*7 +
5)
-
2
tan-1^ + 2).
be thrown
57.
Although the expression px
+ q may
into the form
we by inspection, might proceedthus
:"
Let where
X and
pa?+gsX(2oaj+6)+/i,
/x are
constants
to be determined.
Then
by comparingcoefficients,
pb
=
giving
X
=
and
--
58
INTEGRAL
CALCULUS.
EXAMPLES.
Integrate
1.
f
2
xdx
.
4.
f fo+1)^
//y"
x* + 2x+l 3.
/7/y. ^a^
.
5.
J x2+?
6.
/Jfl-LZ-^p.
/"
f/v"
I
\2
/ 0^t1
"
c^
58. General Fraction and Denominator. of Expressions
are
with
Rational
Numerator
the form
A~4,
9w
where
f(x) and
"/"(#)
functions of x, can be integral algebraic by resolution into Partial Fractions. integrated of putting such an The method into expression Partial Fractions has been discussed in the Differential Calculus forBeginners, Art. 66. When the numerator is of lower degree than the denominator the result consists of the A A
sum
rational
of several such
terms
as
Ax+B
and
Ax+B
the numerator is of as high or higher degreethan the denominator we may divide out until the numerator of the remainingfraction is of lower when in that can degree. The terms of the quotient be integrated fraction and the remaining at once be put into Partial Fractions as indicated above.
A
And
case
may
Now
at
once
fraction of the form any partial into A log(x a).
-
"
-
integrates
A
Any
fraction
of the
1 r"l
form
"
-.
"
ix
^~*
^ a)
into integrates
A
(x"a)r~v
60
and the
INTEGRAL
CALCULUS.
is integral
Ex. 3. Put Hence
Integrate
/
x
^
"
-dx.
\
=
?/.
the fraction becomes until
=
Aj/
remainder,
Dividingout
y3 is a
l+
factor of the
2y
Hence
the fraction
1311111
and
therefore ^2
1 1
and
the
is integral
Ex.4. Let
Integrate
=
a?
1
+y
; then
"We
now
divide
out
by
RATIONAL
ALGEBRAIC
FRACTIONAL
FORMS.
61
is a factor of the remainder. until ty4 coefficients : detached use
2 + 3 + 3 + 1
To
shorten
the
work
we
) 1+2+1
l+
(J |
+
f+f+
i-i-i j+t+ f -f-f-i
i
f+"+f f +tf +if
+
A
tt-A-A
551
e
ll-5y-
Now and
11
-5j/-5y2
=
ll-
5(^-1) -5(^-l)2
by
Rule
Gale, for Beginners, 2, p. 61, of the Diff.
5#2
1
\(x)
and
3
x
+ 1
3(^
3
l+#
3
Thus
ijj
^.2
~
I!
i
-L
+ 1) 2(^7 1)4 4(tf I)3 8(^7 1)2 (a? 1)4(^3
-
^
5
1
1
A7
1
(2a?-l)-3
^2-
^-1)
and the is plainly integral
48
+ 1
6
62
INTEGRAL
CALCULUS.
EXAMPLES.
1.
Integratewith
regard to
x
the
followingexpressions :
v11*'
w'
2
T~\*
\V11V
~f~
\7
?T7
\'
^+ (iii.)
(iv) -/
a)-1^ + b)~\
(viii.)
"a*--")
"
"
"^
2. Evaluate
3.
Integrate
(i) W
dx f J (^2+a^2+62y
(iii) "'
J
f
4.
Integrate
(xd* (i.) v ;
J^+^2
.
[' (iii.) v
(iv.)f
-
+ l
J^+l
cto. Aa?2"t1
do?.
J^4-^2+l
(v.) r (vi.) /"(^-
RATIONAL
ALGEBRAIC
FRACTIONAL
FORMS.
63
5.
Integrate
/.
v
xdx
.
.
x
dx
dx
(vii.)
(iiL"
(^"T4)^"**"
^ ("")
(x\ VA*/
W
~(~"
lX-^-4)' (x*+ i\/
i
6.
Integrate
~3,J~* J~.
(VI.) -7
x"" \
~t
:
d^t?
/T
\o/i
f j
~t
""
\
\
"/
j ^
o\'
V*-'"^*^
(viii.)
''
+ iy (#-l)2(#2
J
*
^2
+
1)3
7. Evaluate
/Vtan~"^(9 and P\/c
the value
8. Obtain
c
o
cos
x
dx
9.
Investigate
10. Show
that
r. fa
"o
_f^
64
INTEGRAL
CALCULUS.
11.
Prove
that
[+*
J
dx
_2?r 2?r
~~
a
+
b
(x*
"ax+
"2X^2
"
bx
+
b'2)
V3
ab(d"
[COLLEGES
7,
1891,]
12.
Show
that
the
sum
of
the
infinite
series
be
can
expressed
in
the
form
and
hence prove
that
[OXFORD,
1887.]
CHAPTER
VI.
SUNDRY
STANDARD
METHODS,
f
doc
60,i.
Integration
of
where
R
-y=
=
ax2+2bx+c.
Case
When
I.
a
Positive.
a
is
positive
If
we
may dx
write
this
integral
as
a
a
which
we
may
arrange dx
as
If
I
__
If
Q p
I
_
dx
.
____.^=._.=_..___^_^==i
__
aJ 7/
+,
"\2
bz-ac
x/"J
according
form of
as
62 is
greater
or
less
than
ac,
and
the
real
the
integral
ax
,
is therefore
(Art. 36)
1
. ,
~
+
b
or
ax
.,
~
+
"
b
,
=
cosh
*
*
smh
"7^
T 1
,
^
Vo2
"
ac
Va
x/ac
62
according
E. T. C.
as
"2 is
"
or
"
66
INTEGRAL
CALCULUS.
In either
case
the
integral may
be written
in the
form logarithmic
~T=
log(ax+
^
"
b+
*Ja*Jax2 + 2bx+c),
~
_
the constant
T=
*J a
Also since and
1
=
logv
cosh sinh
ax
.
~
62
ac
beingomitted,
~
lz lz
=
sinh cosh
1
l\/z2
"
1
, ,
~
~
=
l\/z2 + 1
.
,
cosh
,
~
, l
+ b
=
"=.
sinh
\/aR
"
x
I
.
, "
,
ax
+ b
" "
1
=
,
and
sinh -7=.
V^
-1"
\/ac
7--
"
b2
"7^
cosh
T 1
-
x/aJi
\/ac
7
-?
"
\/a
b2
which and
forms therefore may be taken when a is positive less than ac respectively, b2 is greater or
61.
Case
II.
a
Negative.
f
"
dx
'
If in the
integral
A.
Then 1
r
)*Jax2+2bx+c
.
/
a
be
negative
write
a=
"
our
integral may
dx
be written
ZJ
or
or
"7=:
sin
68
INTEGRAL
CALCULUS.
Ex.
2.
Integrate
(
J
be
dx
This
integralmay
written
I
dx
and
therefore
is
"
=
sin"1-^^"
.
\/2
which may also be
\/41
expressed as
-^cos
V2
-F="
*/41
EXAMPLES.
1.
Integrate
{--^"
JV^
+ 2a? + 3
{
J
-,
dx
2.
Integrate
J
/"
dx
dx
A/2"Ja + Zbx+cx*dx
2"#"
3.* -2#2
3.
Integrate
\
(c positive).
4.
Integrate /\/a +
cyPdx
(cpositive).
62. be
Functions
of
first
the
Form
-.
-"=====
x/a^2+26^+c
integratedby putting Ax+B
into the iorm
may
which,
or
may
be done
as
in Art.
;
we
57, either by inspection
obtain
by equating coefficients
Ax+B
ex/
SUNDRY
STANDARD
METHODS.
69
The
of integral A
the first fraction is
and
that
of the second
has
been
discussed
in Articles
60, 61.
EXAMPLES.
Integrate
-
2.37 + 3
x+b
POWERS 63. Sine Index.
AND
PRODUCTS
OF
SINES
AND
COSINES.
Odd
or
Cosine
with
Positive
Integral
can
of Any odd positive power thus : immediately integrated
"
a
sine
or
cosine
be
To
integrate Isin2n+1# dx, let
.'.
cos
x
=
c,
smxdx=
"dc,
Hence
fsin^+^cfo ((I-c2) dc
=
-
__
70
INTEGRAL
CALCULUS.
Similarly, puttingsince
we
=
s, and
therefore
cosxdx=ds,
have
=
Icos*n+lx dx
(1
"
s2)nds
L_
"" '
nn
I
/
1 \7l.
64. Product
of form
sin^
cos?#, p
or
q odd.
product of the form method admits of immediate by the same integration either p or q is a positive odd integer, whenever whatever Similarly, any
the other be.
For
to integrate/sin5# .example,
-
cos4#
dx, put
cos#=c,
and
therefore Hence
sin xdx=
" "
dc.
"
/cos%
sin5^?dx
/c4(l c2)2dc
cos5^7 cos9^? 9cos7^;
~5~'
J^f~
we
"T"'
:
"
/^
sin5^ cos3# dx
proceedthus
=
I sin^(l sin2x)d (sin x)
-
65. When in terms For p + q
=
p+^
x x
=
is
or
a
negative
even
sin*tocos% expression of tan
admits
cot
x.
of immediate
integer,the integration
=
put tan
"
t,and therefore sec2^ dx
dt}and let
Thus 2n, n beingintegral.
=
|
^
)n
~
ldt
8a5
,
"
4-n~*-(j
Irftan^+6a5
__
I-
4--
p + i
2
SUNDRY
STANDARD
METHODS.
71 cosec2^ dx
if Similarly, and
we
put
cot
x
=
c,
then
"
"
dc,
\"DPxco"xdx=
-
a
result the
order.
Ex.
1.
same
as
the former
arrangedin
the
posite op-
Integratef?^"fo?.
J
sura
This may
be written
-
/
and
the result is therefore
It may
also be
,
in integrated
terms
of tan
x
thus
:
"
CcosPx sPx
J sin6^?
the result Ex.
2.
-r-^-dx
=
I
f
1
"
/T
.
o
2
\,,
an x=-
tan~5#
" -
J taii6
same as
^
"
-
being the
before,
/sec" (9cosec"
0 d"9
=
ftan~*0rftan0"" -f tan~%=
-
f cot*0.
66. Use
of
Multiple Angles.
of a sine or cosine, or Any positive integral power of sines and integral any product of positive powers in be expressed means can cosines, by trigonometrical of the angle, series of sines or cosines of multiples a and then each term be integrated for at once; may f
siunx
nx
7 dx
=
"
J
\cos
and
n
ifsin
nx
dx=
7
cosnx
--.
J
n
72
r
INTEGRAL
CALCULUS.
J}x x.
i
1
.
/ cos
/^^2^^r^,_
x
c/Lx =:
/
f /"-t-UU"^^
X
.
Sin 2.27
Ex.
2,
Ex. 3.
cos%
dx=
/"
_
/
j=
"4"
2# + 2# +
""**
/(|+ J cos
J cos 4dc)dx
"
%x + J sin
sin Ax. g1^
67. It has is odd
no
been already
shown
that when
the index thus in
such
transformation
is necessary,
the second
example
/ cos3# dx
=
/(1 sin2.2?)a? sin x
-
=
sin
sm x
"
x
^
which method
more
presents the
we are now
result
different form. will therefore discussing
in
case
The be of
value especial
nor
for the odd.
of
sin^cos?#, where
neither p
Ex. Let
4.
cos
x
q
are
Integrate I8m9xdx.
+
c
sin
2
x
=y
x
=
;
then
w
cos
2t sin
x
=
y
"
-,
2i sin
nx
"
yn
"
"
y
yn
Thus
=
2
cos
8^-
16
cos
6# + 56 cos4o?-
112
cos
2^+70.
SUNDRY
STANDARD
METHODS.
73
56 2# +
Thus
sin8.??
=
l(cos 8x
2
-
8
cos
6# + 28
cos
4#
-
cos
35),
Q*
andj
\$n$xdx="
J
5.
cos
f
"
K
j
irsiii8#
-
"
-8"
Osin6^
"
+ 28
.
OQsin4#
" -
"
-56
Kcsin2^? +35#
.
" "
"1
,
2i L_
8
o
4
2^
_J
Ex. Put
Integrate
/ sin6a7cos2^ o?^.
; then
.#+t sin ,#=y
=
2
cos
8#
"
8
cos
6^+8
cos
4#+ 6^
8
cos
2^; kx
"
10,
"
and whence
sin6^cos2^="
7J
"cos
'8# + 4
cos
"
4
cos
4
cos
2^7 +5
V,
It is convenient for such examples to remember that the several Coefficients may sets of Binomial be quickly in the scheme :" reproduced following
1 1 121 1331 14641 1 1 1 1
5
68. NOTE.
1
10
15
10 20 35 56
5
1 6 21 56 1
6
15 35
7
8
21 28
7
28
1 8 1
70
etc.,
each number
being formed
above
we
at
once
as
the
sum
the 7th
row
it and have
1 +
the
precedingone.
=
of the Thus in
one
mediatel im-
forming
0+1
=
1,
5=6,
5 + 10
15,
10 +
10=20,
etc.;
74
INTEGRAL
CALGUL
US.
/
and
in
multiplying out
we
such
a
product as
(y
1\6/
" -
)
+ (;*/
-
1
occurring above
and all the work
-
onlyneed
are
the
coefficients of
6 + 15
+ t)2 (1 t)G(l
"
appearingwill be
1
-
coefficients of
(1 t)Q (1
-
20+
5-
15
-
6 +
1,
+
coefficients of coefficients of
each
row
(1-/)6(1 + 0
student
are
1-5+
9-
5 + 9-5
1,
+
of The
4-10 are + 1)* 1-4+ 4+ + t)G(l to the being formed according figures will discover the
reason
4+4-4
same
1,
as
law
before. in which
of this
by
forming per-
the actual
of a+fo multiplication
are
+
cZ2+cfa3+... by l + ",
c,
the several coefficients
a,
a
+
"
6,6 +
c+c?,etc.
the required,
Similarlyif the coefficients in work appearingwould be
1+4 1+3
+
(1+04(1
1,
+ 2 +
O2
were
+ 6 + 4 +
2-2-3-1,
1,
1+2-1-4-1
and the last
row are
the coefficients
are
required.
:
"
The
coefficients here
formed
6-4
=
thus
1-0=1,
4-1=3,
2,
4-6=-2,
etc.
EXAMPLES.
1.
Integrate
odd indices in two ways.
doing those with 2. Integrate
3.
Integrate
ir ir
4. Evaluate
/
*0
ft
r"
r
sin^ctr,
/
0
/ cos5^?c?^,
^0
5.
Integrate sin
that
si 2.# cos2.r,
6. Show
/sin x
7. Show
/" N
sin 2.# sin 3# that
dx=-"\
cos
2#
-
cos ""$
4# +
^\ cos
6#.
(i.)I sm
f
.
7
wia?
cos
w^
a^7
=
"
cos(m+?iV v !
" " "
^
cos(m 72-V y )",
"
"
.
76
and
INTEGRAL
CALCULUS.
generally
dx 2w+2aj
=
-
c
-
"c"
^3
-
s"5
/"2n-fl
~
WC"-
.
.
.
where
c
=
cot
x.
70. Odd
cosecant
can
positiveintegral powers thus : be integrated
" "
of
a
secant
or
By
differentiation
we
have
at
once
d
"
and
(n + l)cosecn+2" n
"
cosec7lo?
=
"
x cosecna -7-(cot
doc
whence
(n +
and
+ (ti
1 ) secw+2^jdx
=
tan
x
secn#
+
^
I se
1 ) cose.cn+2xdx =~coix
f
cosecnx
/
+
n
cosecn#
\
dx
I
Thus
as
sec
x
dx
=
+ ^V logtanf^
and
we
Icosec
may infer at
once
#
c?x the
. .
=
logtan^
,
of integrals
.
sec3#,sec5#, sec7cc,
formulae.
Thus
;
=
cosec3aj, cosec5#, etc.,
in the above I, 3, 5, etc.,
by successively puttingn
/sec3#
dx
=
J tan
x
sec
x
+
^ logtan f
-
+
^V ),
/sec5^?da7= J tan
=
#
sec3^+ 1 /sec3^
J tan
x
sec3#+f tan
^7 sec
x
+
f logtanf
-+-
etc.
SUNDRY
STANDARD
METHODS.
77
"
71. Such
in
formulae
the
as
A
are
called with
"
REDUCTION
formulae, and
student We
will meet
many
others
Chapter
sin^cos^
VII.
consideration
as
postpone till that chapter the of such an expression of the integration been have as except for such cases
alreadyconsidered.
72. Since
is
a a
positive power
of
or a
of
or a
a
secant
or a
cosecant
negative power
of
or a
cosine is
sine, and
positive
of
a now
power secant to
cosine
sine
negative
that
we
power
cosecant
it will appear
are
able
integrate any
or integralpositive
negative power
of a sine, cosine) secant, or
cosecant.
INTEGRAL 73. may For
POWER
OF
TANGENT
OR
COTANGENT.
or
of Any integralpower be readilyintegrated.
tannx
a
tangent
cotangent
dx
=
tann
~
2x(sec?x l)dx
"
=
Itann-2a3c?tan# ltanw"
idM.n~lx
-^
n
"
f, J
tan?l-2#cfe.
1
#,
,
And
since
Itan
#
cfc
==
logsec
and
we
Itan%c dx may
=
(sec2# l)dx
"
=
tan
x
"
x,
tan3#, tan4^, tan5#,etc. integrate
we
Thus
have
/tan3#cfo?=
/tan x(sec2x-l)dx
78
INTEGRAL
CALCULUS.
[
f
2
3
this By continuing
process
we
shall
obtain evidently
_
2?i-l +
2^-3
# + (-l)nff, (-l)w~1tan
and
tan-+^=^^
_
tan^^
t
Similarly
Icoinx dx
=
cotn
~
2
cot71-1^
=
"
--
~
r-
71"1
J
f |COtn-2#CfcE,
whilst
icot^^aj
=
logsin x. I(cosec2^ 1 )dx
"
and and
cot2^ ^ therefore
we
=
=
"
cot
x
"
#
;
may
thus
integrate
etc.
or
cot3#3 cot4#, cot5^,
Hence admits any
integral of a tangent power of immediate integration.
etc. \a+bcosx,
cos x as
cotangent
f
dx
74. We
of Integration
may
write
a
+ b
j, sin2| s2|
-
SUNDRY
STANDARD
METHODS.
79
(a
or
(fa Thus
=
-AgU-j
"
2
6
or
"(1)
CASE
I,
If
a
"
b this becomes
tani
a-
6
/a+6
or
tan
,
?|
2J-
Since
we
may
write
this
as
"
b,
"
1
=5- COS
1-
^
g
+ b
2
80
INTEGRAL
CALCULUS.
1
or COS"
"
+ bcosx
7
*-"
CASE
II.
If
a
"
in the form b,writingthe integral
(K
dian.~
,
(2)
in
placeof
the form
we (1)
have
in this
case
by
Art. 54
UjiAj
"
J.
A J ft i _1_ + bcosx
f*na
sv*
A b
"
n a
"
IT. 1 Ib +
17 a
"
Vf^
,
V6^~tan2
v6 +
a
Ib+
a
x
+
\/b
.
"
a
tan
"=
"log
'
.
\J~b +
rjr"
a
"
v
b
"
a
tan
^
"
By
Art. 33 this may
be written
tanh~:
/62-a2
2 tanh
~
or,
since
lz
=
cosh
~
1
1
"
02'
as
we
may
stillfurther
exhibit the result b
1
"
a. a
-L-^
tan2^
2
nX
:cosh~3
1
b+
1
"
^"-
b+
a
or
SUNDRY
STANDARD
METHODS.
81
We
therefore have
"
b,
x
i.e.
dx
a+
bcosx
er "
b.
or
=
cosh-
Jl?-
a+bcosx
one
but These forms are all equivalent, forms is to be chosen when the formula
of the real is used.
75. The
integralof
"
"
-r
a+
b
may
cos x
be im-
+
c sin
x
mediatelydeduced, for
b cosas-fc and
once smx
=
\/b2-{-c2cos(x tan~V ), b/
"
\
therefore
the proper form of the integral at can be written down in each of the cases or a greater
less than
Ex.
^/5*+c*.
dx
cos x
13 4- 3
in H- 4 sin
x
=f [ J
52
dx
--
(where
tana
=
^)
1
/132
_
13 + 5
:-a) coe(# a)
-
12 1
or
-i/2
.'T
"
a\
$. I. C.
82
INTEGRAL
CALCULUS.
f 76. The
dx
" --
integral I
Ja + 6sm#
,
7
.
may
be
easily deduced
by putting
then
f
"
dx
"
f
=B
dv
Ja + and the therefore
cases course
j". o sin
I-
x
Ja +
"^o
-
,
cos
y
its value
may
be written
down
in both
a^b.
it may be also independently investigated
x as
Of
by
first writinga + b sin
+ 26 sin | + sin2|j a(cos2| |,
cos
or
+ 26 tan cos2^(
a
-
+
a
J. tan2^
The
then integral 2
becomes
and
two
cases
arise
as
before.
77, The treated. dx
integral
I"
,
x
,"
may
be
similarly
f
dx
84
4. Prove
constants
INTEGRAL
CALCULUS.
that,with
involved
certain
limitations
on
the values
of the
J J(a-x)(x-(3)
P
and
=L= /"%=
/
a
-
.-(3
v
____
integrate Integrate
,.
(x
"
a)(/3 x)dx.
"
5.
"}
v
C
dx
'
,
JSC
.
r
"^%
\ (1""
}3(l-s
-
f
("'")
/""" \
.r
Ou27
.
(v-)
/
*
2^2
r
4- cos
a?
+ sin
.77
r
U11'-'
J
cos
a
+
COS.T'
^'''^
\
^^
J"2siii2"9 + 62cos2^'
(vii.)
cos
a
+ cos
x
and
(viii-) prove
o
6.
Integrate(i.)f-
dk
/ (ii.)
(iii.)
J
C?Jt'
____
^ V
a(^
-
b)+
"**
V
6(^ a)
-
f
7.
f Integrate 7 I
Integrate
8.
f- ^
-
-
.
J sm^
+ sm2^
9.
Integrate
J
fcos201ogcos^+shl fa. COS0-S1TL0
1+cosx
10. Interate
SUNDRY
STANDARD
METHODS.
85
11.
Integrate
12.
Integrate /
J VI
sm
x
_dx.
x
+ sin
sec^ cosec
13,
Integrate /
"
dx.
x
J 1+
14.
Integrate /-"f^fl_f_.
J
v a
+
b tan2#
15. Evaluate
fVr^
/ 1 +
o
"
"^'
x
sin
16.
Integrate [****"****"",. J logtan ^7 Integrate
.
17.
Vsin 2(9
18.
IntegrateJ
fcot0-3cot30
19.
Integrate / J Wo?
~
20.
Integrate /"7
J (x si
21.
Integrate f-^ Integrate fA/_
'
22.
*
~
CQS
^
6y
V
cos
+ cos (9(1
+ cos 6")(2
6")
f
23.
Integrate
1
"
" .
.
sin
^?
2
"
sin
a?
24.
Integrate f^ (sin 0+ Integrate f
J
"mg-coeg
cos
25.
INTEGRAL
CALCULUS.
26.
.
Integrate
J
I
sin"1
-
dx.
l+x2
27.
Integrate
J
?
\"
( sin^,
28.
[*"-X-dx,
'
{^^dx,
and prove that
Integrate
J
sin
2#
J
sin
3^
J
sin
4^7
sin"r
5 +
[THIN.
COLL.,
1892,]
CHAPTER
VII.
REDUCTION
FORMULAE.
REDUCTION
FORMULAE.
79.
Many
and
functions reducible whose
occur
whose
or are
integrals
of the
are
not
immediately
forms,
In
some
to
one
other not
standard
integrals algebraic
which
at
directly
may
obtainable. be the
cases,
however,
some
sucft
integrals
formula itself
linearly
connected
by
with
may
integral
mediately im-
of
another
expression,
be
to
either
integrable
than For the
or
any
rate
easier
integrate
original
function. be shown
that
instance
it will
(a2 + #2)^fe
can
be
expressed
in
terms
of
+ #2)^fe,and J(a2
this
latter
itself
in
terms
of
which J(a2 + ar)^cfe,
being
a
standard
form Such
the
integral
connecting
Formulae.
of
I(a2 + #2)^cfc may
algebraical
be
inferred. called
relations
are
Reduction
80. methods
The
student
will
realise been
that
several
reduction instance the
have
already
used.
For
88
INTEGRAL
CALCULUS.
parts of Chapter IV., and It is proposed to consider and in the presentchapter, such formulae more fully of some for the reproduction to give a ready method of the more important,
of Integration method by the formulae A of Art. 70. of xm-lX* 81. On the integration for anything of the form a+bxn. In
where
X
stands
several
a
cases
the
can integration
be
performed
in
directly.
I. If p be
the positive integer, binomial
expandsinto
Next and
s
a
finite series, and each term
/v"
is integrable.
suppose
p fractional
=
-,
r
and
8
beingintegers
positive.
777/
II. Consider Let
.'.
the
case
when
=
"
is
=
a
positive integer.
X bnxn~ldx
a
+ bxn
zs,
=
szs~ldz
zs~l
r-
f
and
J
\x
bn)
and
when
"
is
a
this expression is integer, positive the binomial and
directly integrable by expanding each term. integrating
III. When
"
is
a
the expression negative integer,
(zs-a)~n+'
REDUCTION
FORMULAE.
89
may may
be
put
be
TD
into
then
and the integration fractions, partial proceededwith (Art.58).
r
IV.
may
If
"
H
"
is
:
"
an
we or negative, integerpositive
proceedthus
rn
,
_
-
rn
m-\
and is
--
by
cases
II. and
a
III. this is integrable when
a
-
"
"
either
positive or
b + ax~n=-zs.
777
r
"
substitution
negative integer by the That is, the expression is
when integrable
\-- is
n S
or negative. integral, positive,
Three
cases or
thereforeadmit of integration by simple substitution.
(1) p (2) (3)
"
mediately im-
a
integer. positive
an
integer.
an
777
"
[-p
integer.
Ex.
1.
Integrate (^(c
m=6,
3, and
Here
n
=
"
=an n
integer.
Let
so
that the becomes integral
%x*dx=
2zdz.
Then
90
INTEGRAL
CALCULUS.
Ex. Here
2.
+ x^dx. Integrate/ x*(a?
m
=
", n
=
3, p=b
and
"
+p
'
is
an
integerc
n
The Let then and
is / integral
-3-.
XT
the
becomes integral
9
*
which
=
might be put
6, the
into
process of will be avoided effected (Art.70).
sec
fractions. If, however, z be put partial tions fracinto partial puttingthe expression and the final integration may be quickly
82. Reduction Leta with
any + 6xH
formulae
for
\xm~\a
be
"
=
^C;
then
\xm~lX^dxcan
connected
of the
xm
-
six integrals : following \XP
-
1^
m+n
-
\xm-n-IXPdx,
xm
-
n
-
lX?+ldx,
Xm+n
:
"
~
1
to according
the
rule following A
Let indices whose
P
=
"X+1JTya+1 where
and X
and
JUL are
the smaller
of x
in the respectively
to be connected.
two
expressions
dP
are integrals
Find
-p.
arrange Re-
whose
linear functionof the expressions and to be connected. are Integrate, integrals
this
as
a
the connection
is complete.
92
INTEGRAL
CALCULUS.
Integrating, P=(n
and
+
/"( a?fdx-na? 1) I(x2+
n+I
Putting?i
=
5 and
ft
=
3,
((
J
and Then
3i
6.4
Ex.
3.
Calculate
the value
of
[^x^-^ax-x^dx,
to connect
m
being
a
positive integer. We
shall endeavour
with xm~l*J%ax-x*dxy f l%m"JZax-x'*dx i.e.
(xm~\2a-x*fdx. with (xmJf^(Za-x?dx
to P=^m+1?(2a-^)1r according
Let
the
rule,then
Hence
(m
+
xfdx 2) fxm+^(2a
-
-
xm^(2a^
-
+
(2m + l)a fxm~\2axfdx
-
REDUCTION
FORMULAE.
93
a
.
xm*J?Ltix
o
ra
+ 2
Jo
m
+ 2
o
/la,
. _
xm*j%ax
-
x*dX) and
m
be
a
positive integer,
2ra-l
2
2m-l
.
--
2m
.
-
-3
a
3/ J.m-z
5
. _
"
etc.
m
+
1
m
2m
-1
'
2m-3
m
3
' '
m
"mT2
Now
to find
'
m
+ 1
"*4
3
IQ or
I
fj^ax
"
x^dx, put
"cos
x=a(\
Then
an
0).
dx
=
a
sin ( sin 0.
we we
d
^l^ax
"
x^
"
a
Also
when when
#=0,
#
=
have have
$=0,
O
=
2a,
TT.
Hence
70=
fVsinW^-
T(l
-
cos
20)rf0
Hence
/
-m-
1)...3
+2?r_ 2
(m+2)(m
+
l)...3
(2m + l)! m!(m + 2)!
EXAMPLES,
Apply
reduction
1.
the
rule formulae
stated
(when
in Art. 82 to X=a + bxn) :
"
obtain
the
following
/
J
2.
94
INTEGRAL
CALCULUS
3.
(,-^
J
4.
( J
=xm^ (a"*-*X*dx
J
^P
-
.
{x"+n
mm]
6.
.
/
Integrateout
7. Obtain
m
=
the
m
=
of integrals
/xm^(^Lax x^dx for
"
the
cases
l, m=2,
3, and
0 and
their numerical
2a.
values
when
the limits
of
are integration
83. Reduction
A
formulae be
for
sin^a? co"x
a
dx. formula
similar rule may
givenfor
cosqx
reduction
for
This
Isiupx
j
dx, with
any
be expression may six integrals : following
"
connected
of the
I sin^
"
2# cos?#
dx,
\ si
\ si
sin^+2^ cos^
-
\ sinpx
cos9'
~
^x dx,
I sin^
by
the
-
2x cosv+2x dx,
*x
dx,
Put smaller
two
rule. following sinX+1#cosAA+1" P where
=
X
and
^
are
the
indices of since and
cos#
in respectively
whose expressions
-T-,
are integrals
the to be connected.
Find
and
rearrange
as
a
linear
functionof the
ax
whose expressions
are integrals
to be connected.
REDUCTION
FORMULAE.
95
and Integrate
Ex. Connect the
the connection
is effected.
integrals
/"
Let P=s
=(p (p
=
"
"
cosg^(l sin2#) (q-f l)si l)smp~2x cos9# I )sin^~2# ( p -h ^)sin^ cosPx
"
"
"
[Note the
last two
lines of rearrangement and
as
a
linear
functionof
sin^cos^
* . .
sin^~2^7 cos%], dx
P=
(p
-
I ) /siii^~2^7 cos9^
-
(p
+
q)Isi
Hence
/sin.^
cos%^
8m
=
-
*~^X cosq+l* p + q
+"zi (*
P + qJ
It will be where
remembered,
or
however, that
odd
in
the
case
either p
q
is
an
be effected can integration The present method is useful integers. q are both even
integerthe complete 64, 67]. immediately[Arts.
in the
case
where
p and
EXAMPLES.
Connect the
integral fsin*4?cos?#e"i? with
1
.
/
smp+2x
cosq.v dx.
2.
/siii^
I smpx
cos"~23? dx.
3.
4.
/*sin^/ sin^
+
5.
2# cos"~2# dx.
96
INTEGRAL
CALCULUS.
6. Prove
that
fsin^^
_
cousin-1*
n
"-I
n
f^,.
J
J
Employ
7. Establish
this formula formula
to
sin8^. sin%, sin6#, integrate for
a
of reduction
/cosw#
dx,
8.
Integratesin4
84. To
calculate the
V
integrals
71
5^n
=
f 2"
I smn#?
.
aa?
and
J
0
(7n
=
fl I
J
0
Connect
Let P
Isinn^cdx
sinn~3#cos;E dP
"
with
Isinn
~
2x dx.
=
to according
the rule; then
sunx
"
"
"
dx
=
(n
---
"
l )sinn2x
~
"
.*.
f lsinn^a^= J
.
smn~lxcosx
,
"
n
"
If
.
n
n
-\smn~zxdx. J
when
?"
r
Hence
since
sin" -^
cos
"
vanishes
x
=
is
an
integernot
x
=
less than
2, when
0, and
also when
J,
we
have
71"1
=
---
^
"
3
^
---
71
7i
"
5
4
^
*
""wto
a
71
71"2
"
if
71
be
even
this
Ti-l
comes ultimately
Ti-3
5
3 4
Iff
"~'ii^V"g
2j J
REDUCTION
FORMULAE.
97
Ti-1
7i-3 n-2
'"
3
1
that is
n
TT
422'
If
n
be odd
"
we
similarly get
1
"
Q
ll/
^^
A,
TJ
*
.v
f-j
" "
il/
^^
-L
"
O
"r"
9 Z"
I
/" 2"
.
Ufl
.
71
n
"
2
*
*
"P
5
3J
x r
"S I
I
dm w E*-1-11 w
rl w U**/j
-|
and
since
I sin xdx
o
=
\
"
cos
=
1
Ti-1
we
7i-3
1
"
4 2'"
2
have
$n
=
-
5* 3*
be
seen
In
a
similar way
it may
that
I"cosnxclxhas
o
the precisely
in each value as the above integral This may be shown too from odd, n even. case, n other considerations. These formulae are useful to write down quickly of the above form. any integral
same
/""".""",, "^|."
[The student should notice that these are easily by beginning with the denominator. of natural numbers ordinarysequence Thus the first of these examplesis
(10 under
written
down
most
We then have the written backwards.
9)x (8 under 7)x (6 under 5),etc.,
factor
a
-.
and writing at (2 under 1), a stopping first denominator with
E. I. C.
But
when
the
is
odd,
in
forming such
no
sequence
it terminates
(3 under
2) and
factor
-
is
written.]
35 G
98
INTEGRAL
CALCULUS.
r"
85.
To
investigate integralbe
a
formula
for
0
2si
Let this
denoted
by f(p,q) ;
then
since
tanP^^
J
we
p + q
p + qJ p be not
have, if p and
2
q be
and positive integers,
less than
GASP:
I.
If "" 6e
ei"e7i
=
2m, and
# afeo
even
=
2n,
(2m-l)(2m-3)
2rv
/(
)
=
3)...l
"/v
m
2
v
andj
/(O,2")
"
^/^
\
9""
=
/i
7/1
^i
"
1
2ft
"
3
1
TT
0
Thus
CASE
II.
If p
6e
6^67^
=2m,
and
q odd
=
=277
etc.
"
1,
2m~1
/(2m, 2^-1)=
-
-/(2m-2, 2^-1)
'
and
100
INTEGRAL
CALCULUS.
These
relations
will
be
found
an
T(n + 1) where
n-{- 1 is either
sufficientlydefine integer or of the form
to
'
2k
+ 2
1
k
being
For
a
positive integer.
instance,
T(6)
=5F(5)=
=
5
.
4F(4)
=
5
.
4
.
3r(3)
=
5 .4.3.
2F(2)
5.4.3.2.ir(l) )= S
.
=
5!
V-) =F(f
PXiHf
.
I- fr(f )= i
.
.
|
.
f
.
f r(" )
This do
not
function propose
is
to
called
enter
a
Gamma its
function, but
we
into
properties
further
here. The
products
1.3.5...
2n-I 2u
...
2.4.6
TT
which
occur
in
the
foregoing
cases
of
I
o
sin^0
cos?0
d9
may
^
be
expressed
^(2n+l\_2n-l
1
at
once
in
2n-3
terms
2n-5
of
this
function.
lr/l\
'
\~~2~)
2
2
2~
2
V2/'
so
that
/y
7T
2
and
sothat
Hence
in
Case
I.
REDUCTION
FORMULAE.
101
In Case
II.
In Case III.
In Case IV.
we
have evidently
the
same
result.
It will be noticed therefore have the same result, viz.,
7T
that in every
case
we
f
and that the ^
?9 1
+1
+ l
in occurring
the denominator
is
the
sum
of the
is
a
and
the
#4-1
^"
in the numerator.
convenient of the above quickly integrals
very
IT
This
formula form.
for
evaluating
Thus
rs f \in"6" cos80 dO
.,
=
-*
5?T
V^TTj" f f i| ^/?T __f f j" ' " " * "
_
2.7-.6.5.4.3.2.f~~215'
102
INTEGRAL
CALCULUS.
87. The student been pointed out
should,however, observe (asit has
that when either p or q previously), both of them odd the or are expression integers, without reduction sinP$cos?# is directly a integrable formula
For
at all.
instance,
=
J
[sin^(l-sin26'Xsm6'=^7 (sin66"cosW6" 79 J
and
Similarly,
-2 0 cos26"(l cos2"9+cos46")dcos
"i
COS76n" COS3"9, QCOS5"9
-
+2"
3
"Jr4-"-*^*.jo,!
s
But
when
p
and
q
are
both
or integral required,
if the
must
and limits of
even
the indefinite be integration the reduction
other than formula
0 and
^,
tt
we
either
use
of Art. 83
or
as proceed
in Art. 67.
EXAMPLES.
Write down the values of
REDUCTION
FORMULAE.
103
prove
the formulae
(1}
I sm2mOcos2n6d0
J
="
-Em+n
I?L_?.-_. 2
(2)
f sin
J
4.
-Bm+n-l
down
cos
Write
the indefinite
of integrals
0 dO, fsitfO
cos3"9 dO, fsitfO
fsitfO cos50 dO,
fsin70 cos20 d09
Evaluate
cos4^ dO. fsiu60
__"
0
/"
7T
rf
rT
sin5^cos2^^.
J
0
/ sin4#"w,
J
0
/ sm26
/3"
7T 7T
6.
/"
0
/-^
/"
"T
J
the formulae of Art.
84
'
J
for
7. Deduce
/
(
sm x
dx from
the
result
r("ii)r(z"l) y 7
V
27
V
"
f
of Art. 86.
EXAMPLES.
1. Prove
that
=
(a) I cos2w"" ^
"/
-1 tan "/" + cos-nc/"
^^t
M
\
-
~
\f cos2
2iii/ J
(b)
104
2.
INTEGRAL
a formula Investigate
CALCULUS.
of reduction
to applicable
when
m
and are n if ra=5, 7i
and positive integers,
=
completethe
tegrati in-
7.
[ST.JOHN'S COLL., GAME., 1881.]
of reduction for
3.
a formula Investigate
and
by
means
of this
show integral
that
._J_
271+2
2 27i+ 4 2.4 271+ 6
adinf
2.4.6
271 + 8
2. 4. 6. ..27i
~~3. 5.7...27i + l'
Sum
1
I
m
,
also the series
1 2 1
I
t
1.3 2.4
1
I
1.3.5
"
1
_1_
,
.
OjCll ITlrT
\f
271+1
4. Prove
2^ + 3
27i + 5
2.4.6
271 + 7
[MATH. TRIPOS, 1879.]
that
2n+l
/
(rf
*-,
6. Find
prove
reduction
formulae
for
+ bx)P*dx, (a) x"(a
J
(y)
and
obtain the value of
*^*"+a")"*r, (S) /2p+l /*"(*"
-
.
[COLLEGES"CAMR]
n
7. Find
a
reduction
formula
for
Ieaxcosnx dx, where
is
a
and positive integer,
evaluate
[OXFORD, 1889.]
RED
UCTION
FORMULAE,
1 05
8. Find
formulae
of reduction
x
for
/#wsin
Deduce from the latter
dx
a
and
/eaxsinnx
of reduction
dx. for
formula
a*
Jcos
Tt
8in"*"fe.
[COL1KGES
%
189o.]
9. If
un=
rT / si
o
prove
that
^-lrc--
and
deduce
un=
-"
2n+1 In
-^^-+^
n("" 1)
rC-f1-/ /b
--
--3),
sv
+
\
""
"(^" l)(n-2)
'
f
(2ro-lX2ft-3)...3 TT
8'
[MATH. TRIPOS, 1878.]
10.
Show
that
1 V
1 \
/
'
wi"
2/3
n"2fifJ")
[TEIN. COLL., CAMB., 1889.]
11. Prove that
7
*
1
1 2.4.6 ...2m _1.3.5 ...(2w-l) TT ~2.4.6...2m 4~3.5.7...(2m+ l) 2*
'
12. Find
a
formula
of reduction
for
f-~=L
^
v'^
"
Show
that
1
3. 5. 7.
..(27i+l)l
where
13.
a1} a2,
...
are
the binomial
coefficients.
[ST.JOHN'S, 1886.]
Show
that
mx
2TO
/cos
=
cosm#
dx
sn^ mm
"
r+-
4~~
sn^
~"
,
~T72
where
m
is
an
integer.
[COLLEGES
a,
1885.]
106
14. Show
INTEGRAL
CALCULUS.
that
m
being a positive integer.
15. Prove
[OXFORD, 1889.]
that if
Im,n=
(m + n)Im) n=
"
I cosm#
cosmx
1
sin
cos
nx
da:,
+
nx
m/m_i? n_i?
+
[if ^-l^(2+i+i
16.
/
92
93
-+^J- [BERTRAM]
m(m"l)T v
m*-n2
9"i\
If
/m?
/m
r
=
w
I cosw#
cos
nx
dx,
prove and
that
,1
cos%^7
w=
"
-
,
m2-^2
^
^
"I
d f cosmx\
-
cEr\oos9u;/
)+
./^m-2, n,
show
that
/I
prove
cosm^7 sin
7^^;
dx\
that
^m
'
w=
--
1
"
"
wm_i
+ n
M_I.
m
+
?i
m
Hence
find the value
(when
cosm#
m
is
a
of integer) positive
/If
o
sin 2mx
dx.
[7,1887.]
18. Prove
that
J
19. If
m
' r2 / cosnx
cos
nx
dx=
IT
"
2n+1" that
T
[BKRTRAND.]
+
n
be even,
prove
/
co
-
m-n.
i
. -
1
2
2
[COLLEGES, 1882.]
108
INTEGRAL
CALCULUS.
28.
Find
a
reduction
formula xmdx
'
for
the
integral
(log#)n'
29.
[OXFORD,
1889.]
Find
a
reduction
formula
C
for xmdx
[ft 1891.]
30.
Prove
that
if
X=a
Z""=/n
[ST.
31.
JOHN'S,
1889.]
Find
reduction
formulae
for
\CLJ
I tann
x
dx.
(Q\
f
J (a+"cos#
dx
.
+
csin#)n
32.
Establish
u
the
v
following
functions
formula of
x"
for and with
double dashes
integration denoting
to
x :
"
by
parts,
and and
being
suffixes
entiation differ-
integrations
respect
/ /u
(-
+
l)n-1nu^n-1hn+i
+
(-
l)"n I uMvn+idx
+
(- l)n
{dx (u^vndx.
[a, 1888.]
CHAPTER
VIII.
MISCELLANEOUS
METHODS
AND
EXAMPLES.
f
INTEGRALS
OF
dx
."
FORM
\^
of
88.
The
integration
of
expressions
dx
the
form
can
be
readily
I.
effected
in
all
cases
for
which of
X X X
and
Y
are
both
linear
functions
x.
II. III.
linear,
Y
quadratic.
Y
quadratic,Y
linear.
If
X
and
be but
both
the
quadratic
process is
more
the
integration
troublesome.
can
be
performed,
89. The
CASE
I.
X
and
Y is
both
linear.
best
substitution
:
"
Let
-fe
dx
110
INTEGRAL
CALCULUS.
Putting
,
cdx
,
we
nave
=
,_
at/.
and
ax
+ b
=
-(y2
C
"
e)+ b,
and
/
becomes
21
"
"
jay2
the standard forms
^
"
ae
7 + bc
.
,
which, being
one
of
Jy
Ex.
2_^ "A
is
2,
immediately integrable.
Integrate /=
f
"
J (xLet
then
Thus
y-l
y+lj
90. The for the
same
viz., substitution, *Jy=y will suffice
l(fi( T\f] 'IT
of integration
I
-^ ^
when
is ^(cc) and X
any
rational F
are
functionof x, algebraic integral
and
eacfe linear.
Integrate /==
Ex.
.
J^have
f
Writing "/^+ 2=7/, we
%dy
and
.r
=
?/2 2,
-
MISCELLANEOUS
METHODS
AND
EXAMPLES.
Ill
so
that
-"L ="="
+
24/-32/
+ 16
(by common
Thus
division).
91. CASE The Put
II.
X
linear,F quadratic.
"
proper
substitution is :
X=\
y
Let
Putting
we
ax
+ b
=
-
,
t/
differentiation, have, by logarithmic
adx
ax
dy
y
+ b
and
ex2 +
ex
+/=
6Y + ft) -2((+/ a\ a2\/
-
-
/
/
Hence form which
the
has integral /=
:
"
been
reduced
to the
known
has been
alreadydiscussed.
112
INTEGRAL
CALCULUS.
Ex.
Integrate /= #+l=y-i,
/= then
f
./
__=_
Let
and y
#+1
__
i+i-2
i+%-y2
JI*
V
92, It will form be
now
appear
that any
of expression
the
J(
can
f
rational
common
being any integrated, "j)(x) function of x. For by algebraic
"b(x)
we can
.
integral
division
express
-
,
in
the
,,
"
torm
Af
Axn+Bxn~l+
remainder.
...
+Z
being the quotientand
M
the
We thus have reduced the process of a number of terms of the class integration Eaf
to the
and
one
of the class
f
M -dx.
The
and latter has been discussed in the last article, of the former class may be obtained by the integrals formula
e
v '
reduction
^(^_^/)4_2r-l
TO
r-lf
v
2r
c
r
c
MISCELLANEOUS
METHODS
AND
EXAMPLES.
113
where The
Ex.
F(r)stands
for I
J
f
,
xr
dx.
as
an
\Jcx*+ ex+f
exercise.
proofof
this is left
Integrate /=
division
f ^2 + 3^+5-^.
J
(x+l)*/x2+l
*=x
By
Now
*2+3*+5
+ 2 +
-
and
to
integrate /
x we
-
put #+!=_
and
get
Thus
93. CASE The Put Let
III.
X
quadratic,F linear.
"
proper
substitution is :
+/Y=y.
T /=
f
I
-
dx
"
J (ax2 + bx + c) V
j
"c
+/
Putting
*Jex+f=y,
edx
7
and ax24-"^ +
c
reduces to the form
E. I. C.
114
INTEGRAL
CALCULUS.
and
I becomes
-f
can
2
i
dy
be thrown into
e)Ay*
fractions partial
Now
as
"
-j
4
.
p
n
and
each
fraction
is
by foregoingrules. integrable
that the
same
94. It
may
is also
evident
substitution the
be made
for the
of of expressions integration
form
f
_
*(")
__
dx"
J
where when
is "p(x)
c)\/ex +f rational,integraland
(ax2+
bx +
to yt y"
_i_ "L
"
algebraic ;
"
for
to
^/ex+ fis put equal
\
reduces
0 2
7
the
and
as
form the
"^
/j/2nj_ \
/,/2n-2_i
which
'
-
by divisi"n'
be
rules
for
partialfractions, may
expressed
and
each
term
is at
once
integrable.
Ex.
Integrate
/=
Putting \/^+ l=y,
we
have
-7====2e?y,and
v^
+
1
_
2
116
INTEGRAL
CALCULUS.
Thus Also
so
/becomes
(a2
"
W
that
and
Thus
/ reduces
further
to
If
a
"
6,we
may
arrange
/
i
-
as
/"
V^TP
,
Ex.
.
2.
/= Integrate
J (2x2 \/3^2-2^ 2x2 -2^+1) -2^
__
/
-
+ 1
-,
,
.
1
dy
_
3^-1
~~
2#
"
1
values yj2and yf of ;/2 and minimum are given 2 by x \ and # 0, and are respectively and 1, so that for real be not greater than 2 and not less than 1. values of #, ?/2 must The
maximum
= =
MISCELLANEOUS
METHODS
AND
EXAMPLES.
117
Now
yi-f=^-y"='
2^2
and Thus / becomes
t-tfm^l-g
/2#"
-
_
2ar+l )(2a72 r("g3J x(x-l)
2.37+1
Now
~
1
Thus
'=/(-,==
2
=
cosh"1?/ +
2
cos"1-^/3^2-
,
1
N/2 \2^-
EXAMPLES.
Integrate
1. 4.
_4
"
2.
5.
"
3.
6.
118
INTEGRAL
CALCULUS.
96. Fractions This fraction
A
of form
can
al
ina;+CCOSa;. CT+f + Pisin^ + CfiOSX
into the form
"
be thrown
1
1
~
+ ^sinx (Oj
+ c1cos x)
are
x + + b^siu (ax so
where
A, B, G
A + term
constants
chosen
c-fos x) that
Ca^a,
and
each
-Bc^ + Cb^b, is then integrable.
97.
the expression Similarly
a
+ b sin x-\- c
cos
x
may
be
arrangedas
cos
(a
+6^+0
.)"+
_|
-
x + + 6xsin (Oj
x)n c-[Cos
~
the first and reduction formula and
third fractions may
be reduced
by
a
while [Ex. 25, Ch. VII.], is immediately integrable.
the second
98. Similar
a
remarks
x
apply to
x a
fractions of the form
x
+ b sinh
+
c
cosh
+ b sinh
+
c
cosh
x
ai + 99.
#' (ax # + qcosh x)n' x + Cjcosh + 61sinh frisinh
Some
Special Forms.
that sin
a?
It is easy to show
Isin^r sin
"
"
c\
a
.
"
'sin(a 6)sin(a c)
"
a),
MISCELLANEOUS
METHODS
AND
EXAMPLES.
119
sin2,^
and
" "
7
r"
-.
"
"
-.
"
a)sin(^ 6)sm(#
""
T\".
-,
"
r
c)
1
" ^^ "
sin2a
1/ ^^i E51111
1
,
/
__ ^^
Iv
r)\"nY\( rt (-t/ U lollll
/^ I/ I
QTTn
ollll
/7* cC"
" ^^
/Y
i tv^
f
whence I
-r",
sin
.
#
c?a?
M
"
/
.
\
Ssina 6)sm(a sm(a (a 6)sm(a
-^"7
,
.
.
"
7
. v
.
, /
r
"
" "
c)
r
a), lo^sm(o3 v 6
and
,
f
--
sin2^ dx
-
J sin(^ a)sm(x
"
-5-7
r-^-p
"
/ x o)sm(x
,
.
r
"
c)
x
"
sin2a
a
"
S
100. More
-
"
sin
(a
;
-
"
6)sm(a c)
"
1\
"
/
:
tan
"
7r
2
.
integrate any
_
Hermite generally of the expression
has
shown
*
how
to
form
where
Ti
"
8,cos 9) /(sin sin($ a1)sin(0a2) sin(0 an)' f(x, y) is any homogeneousfunction of
_ "
"
"
.
.
.
#, y
of
1 dimensions.
For
by
-
the
rules ordinary
_
f(t, 1) (* Oj)(t a2) (t a^
-
.
.
.
fractions partial 1) /(a,, (ax egCoj a3) (ax "") of
_
-
-
-
.
.
.
x
,
^" ax
(a2 ^Xag
"
"
a3)
...
(a2 an) ^"
"
a2
which
may
be written
__
_
^r((ar a1)(ar a2)
" "
...
(ar" aw) ^
"
ar
(the factor
of the above
ar
being omitted coefficient).
"
ar
in the denominator
*
Proc.
Lond.
Math.
Soc.,1872.
120
INTEGRAL
CALCULUS.
Putting
theorem
"
=
tan$, a1
=
tana1, a2
=
tana2, etc.,this
becomes
0, cos 6) /(sin sin($ a1)sin(0a2) sin(0 an)
"
"
"
.
.
.
/(sin ar, cos ar) r=isin(ar OL) sin(ar an) sin($
" " "
...
ar)
Thus
W: /(sin0,
~ "
cos
9)
/(sin or,
"
cos
"
a,)
7
\
"
"
^ism(ar
x
./
7
^
"
"
%)
.
.
.
sm(ar
an)
_logtan^ 2
J-'-'ii t/ctj-j.
;-:
.
EXAMPLES.
Integrate
sm ^ cos
^^7
x
-
4
cos
"
cos
a
cos cos
2#
^
"
cos cos
2a
a
K
sin 2^7
sn ^7
"
sin 2a
sn
a
"
"
0
cos
"
3#
x
"
cos cos
3a
.
O.
D.
cos
"
a
sin
sin2a)' #(sin2,#
"
GENERAL 101. There
are
PROPOSITIONS.
on general propositions
certain
almost which are integration definition of integration or meaning. Thus 102. I
self evident from the from the geometrical
f(j)(x)dx= J
~~
for each is equal to \^(^) \HC0 if "f"(x) he the differential The result beingultimately coefficient of \fs(x).
*
See Hobson's
Trigonometry, page
111.
MISCELLANEOUS
METHODS
AND
EXAMPLES.
121
of independent
z
x
it is
is used
in the
immaterial whether x or plainly the indefinite process of obtaining
pc
/""
integral.
/""
103. II. For if
1 "p(x)dx "f)(x)dx + (j
=
a
a
c
and which Let
of "p(x) integral the left side is \{s(b) side is \^(c) the right is the same thing. illustrate this fact geometrically. us be the indefinite \[s(x)
"
Let dinates
the
curve
drawn
N^^
Then
NJP^
be 2/ N^P^ be cc
=
anc^ 0(#0"
a,
x
=
=
c,
x
=
^e^ ^ne or" b respectively.
the above
equationexpresses
+
area
the obvious
fact that Area
104. III. For with the
["j)(x)dx [$(x)dx.
=
"
a
b
same
notation hand
as
before
"
the left
side is side
and
the
righthand
\/r(6)T// is [
"
122
INTEGRAL
CALCULUS.
105. IV.
f0(a x)dx f"/"(x)dx
=
-
0
0
For if
we
we
put
x
=
a
"
"
y,
have if
dx
x
=
dy,
y
=
and
=
a,
0,
Hence
I "p(x)dx= I (f"(ay)dy
"
"
o
a
=
(by in.) fV"-2/X2/
o
=
I "{"(ax)dx (by I).
"
o
in
this expresses Geometrically 00' QP the area estimating
the
obvious
between
fact that, the y and x
O'
Fig.9.
axes,
an
ordinate
our
like take
as our
O'Q,and a curve at 0',O'Q as origin
PQ, we may if we our F-axis,and O'X
direction positive
of the X-axis.
124
Thus since
INTEGRAL
CALCULUS.
sin"^^
sm"(7r #),
-
/
0
smnx
dx
=
2
/ sinw^7dx
0
;
and and
since
cos2n+1# cos2n#
rir
=
=
cos2n+1(?r x\ x\ cos2n(7r
" " "
I *o
+ 137(jfo7 COS2'l 0,
=
and
/ cos2w# dx
'0
r
=
%\
ft
dx. cos2n^7
0
We To
may put such a add up all terms
0 and
TT
into words, thus : proposition of the form smnxdx at equal intervals
"
between
to double.
is to add
up
all such
terms
from
0
to
"
and
For the second quadrant sines are merely repetitions order. of the first quadrant sines in the reverse Or geometrically, the curve about $mnx the ordinate being symmetrical y
= =
#
the ^, and
whole
area
between
0 and
TT
is double
that
between
0
|.
illustrations geometrical will
Similar
apply to
other
cases.
108. VII.
If
/"net
""ti
\
For, drawing the
"j)(x)dx=n\ "j
curve
pa
it is clear that it "j"(x), of the part consists of an infinite series of repetitions the ordinates OP0 (x 0) and JV^Pj lying between bounded (x a} and the areas by the successive of the curve, the corresponding ordinates and portions the #-axis are all equal.
y
= =
=
Thus
f "{"(x)dx=r'(t"(x)dx= f )dx
j"wa
I
/"a =
=
etc.
and
I
71 1 "p(x)dx. ^"(x)aa;
I
MISCELLANEOUS
METHODS
AND
EXAMPLES.
125
Thus, for instance, f2"
J
"
sin
xdx=%
o
J
\
F
"
sm
"",
"
j
\276u?
=4
A
J
/
T
-
Bin
"n
7 #aa?""*4
A%n-I
""
2n-3
...-"
I
2
IT
-.
2?^
2ra- 2
2
O
Fig. 10.
SOME 109. We
ELEMENTARY have
seen
DEFINITE
that
INTEGRALS.
the
whenever
indefinite
value
the be performed, can l^"(#)cfe integration the
of
definite
can integral "j)(x)dx
at
once
be inferred.
the
In
many
cases,
however,
the
value
of
definite
definit be inferred without can performingthe inintegral when it cannot and be even integration, performed. We propose to give a few elementaryillustrations.
Ex.
1. Evaluate
/=
=
{*(
J
Writing
we
have
and
vers~
-
=
TT
"
A
a
a
126
INTEGRAL
CALCULUS.
Hence
1=
-
~3/2f(TTf\2ay
Hence
/=
|
o
Putting
and
we
?/=
a(l-cos0),
obtain
/=?an+1
f'smn+10d0
=
7ra
n+i
...
down
to
? or
3
l 22
E,
as according
n
is
even
or
odd.
ir
Ex.
2.
Evaluate
/=
/ logsin #
0
ofo?,
Let then and /=
#=--#, 2
dx
"
"
dy
;
"
/ logcos y dy
rl
=
/ logcos
#
c"
rf
Hence
2/=
/
\
o
logsm^"ir+ / logcos
xdx
jo
log /I (log /f
o
IT
sin
^
cos
#
c?^
sin 2%
"
log
r"
"j
0
Io88inte"fo-i
2x=z,
o?^7
=
Put then then
^dz ;
I f
/
Iogsin2^^=^/logsin zdz"
MISCELLANEOUS
METHODS
AND
EXAMPLES.
127
Thus
27=
log 2, /-^ 2i
/=|logl.
log /?
r~% -\
sin xdx"
\ log cos
J
#
cfo?
=
-
log
-.
2t
2i
Ex.
3.
Evaluate
1=
/ -^
o
Expanding
the
we logarithm,
have
6
If
we
put
/=
"
x=l
"y, /
"
re
have
/
"
^-dy
=
dx.
Hence
we
also have e
J
1
/
"^'a?^=
\-x
"
"
.
6
"o
Ex.
4.
Evaluate
-I
log(tan0+cot0X0
o
Put
^=tan(9,
.-.
1=1
(log /2
o
IT
sin ^ 4-
log cos $)6"
-
2
sin /log
0 dO
=
Tr
log 2.
128
INTEGRAL
CALCULUS.
110.
Differentiation
under
an
Integral Sign.
to be "p(x, Suppose the function to be integrated c) which is of a quantityc independent x. containing Suppose also that the limits a and b of the integration and of are finite quantities, independent c.
Then
will
-
J0(a?,
a,
r"
For
let
u
=
f6 \ "f"(x, c)dx.
a
Then
u
+ Su
=
f 0(o3,
"
which, by Taylor's theorem,
And be
if z, say, be the
value greatest
of which
capable,
in the limit when vanishes and in the limit Thus diminished. ^
"
Sc is
indefinitely
'
=
"
'dx.
a
MISCELLANEOUS
METHODS
AND
EXAMPLES.
129
111. contain
The
c
case
in
which
is somewhat
the limits a and beyond the scope of the
b
also
present
volume.
112.
new
be used to deduce many proposition may has been performed. when one integrations This
since
Thus
f
--
L=
=dx
=
-*
Vc
tan-1
+
a
\l2=2(c+
*
a"
0),
c,
J
(x+cyJx-a
c
+
a
we
times n have, by differentiating
with
regard to
times with regard to n Also, differentiating
a,
we
obtain
/"
(^+ c)(^ a)
c,
r we
^
2
this differentiating Similarly, obtain o?^?
latter p
times
with
regard to
IJ
"2^+1
(^+c)^+1(^-a) 2
EXAMPLES.
1. Obtain
the
: integrals following
"
f(i+*)-V*"fo. (v.)Jf (i.)
J
An-^-xi+ (ii.)
J
ar)-*^
(vi.)
J
I
*?)"*"". r#-1(2-3a?+ -(vii. (iii.)
E. T. C.
130
INTEGRA
L
CA LCUL
US.
2.
Integrate (i.) + (a2
1
62
-
^2)v/(a2 ^2)(^2 62)'
-
[ST.JOHN'S, 1888.]
(x2 +
a2)^^+^
I~STJ"HN'S" 1889.]
|UL*
sin
3. Find
/-
6"Vacos2^ + 67iii2"9+V
sin dx
x
[TRINITY,1888.]
the values ^
f
of
x
J (cos x
/"
+
cos
a)V(cos x 4- cos fi)(cos+
cLx
cos
y
)
C% 1890-]
\
I
J cotfx + a\Jcos(a;+B)coa(a: + 'v} a)\/cos(^ + ^)cos(^ + y}
4.
Prove
constants
that,with involved,
certain
limitations
on
the
values
of the
d,L
\^ Olll
"
"
.
+ Zbx + (x-p)(ax?
cy*
(-ap2-2bp-cft
be
(x p)(b2 acf
-
-
[TRINITY,188G ]
5. Prove
that
\(cQ$x}ndx may
expressed by
j\r ^.v3
"
the series
_L
-"
-r
...
pf p etu,
n-
ND N2J Nft and n having any
. . .
being the coefficients of the
real value
expansion(1+ a) 2
,
or negative. positive
[SMITH'S PRIZE, 1876.]
6. Evaluate
the
W
definite integrals : following
"
(i\
J
fl
l
/a(a2
o
/"""
^2
/y.2
^'
+
^72)2
** x
[ST.JOHN'S, 1888.]
^UL'
\
f
dx
+
(l+tfX2+^)(3 Jo
#)
[OXFORD, 1888.]
7. Prove
that
f
-
8. Show
that
[0x^,1888.]
132
INTEGRAL
CALCULUS.
15. Evaluate
(i.)f
o
dx
[I.C. S., 1887.]
[I.C. S.,1891.]
16. Prove
(i.)
J
f^an^^
sec
37+
cos
#
4
[POISSON.]
J a2
a
"
cos2.2
beingsupposedgreater than unity.
17. Prove
[OXFORD, 1890.]
(i.)
f 1-2S"fc? g
=
-
o
18. Prove
that
=
a
"
"a3
3
+
""
"5
"
" " -
ctf +
...
z2)
19. Prove
IT
3.5
3.5.7
[OXFORD, 1889.]
that
/-7A "^"
_T
2r~% /
.
2 1 1
1 2
o
"
1*3-4
Q2
1
i
1
2
"
o2
^"
"
PL2 " x,6
i
beingsupposed "
20.
1.
Prove
that
[MATH. TRIPOS, 1878.]
21. Prove
that
1.1
1
''*
[A 1888.]
22.
If
$(x)dx= -,"),\*a
-
F $(x)dx.
0
[TRIN. HALL, etc.,1886.]
23. Prove
that
b
*""6^ Jf^C~^W
c-a?) ^-6)
x
provided
[ST.JOHN'S, 1883.]
remains
finite when
vanishes.
MISCELLANEOUS
METHODS
AND
EXAMPLES.
133
24.
Prove
that
and illus/""{"##) ra$(x)dx= + ""(2a-a?)}cfci7,
*
trate
the theorem
geometrically.
show that
25. If
f(x)=f(a+x\
and
illustrate Show
geometrically.
q-pj
\
26.
that q-p q-p
sums
value of the the limiting 27. Determine by integration of the following series when n is indefinitely great :
"
'
n
+ I
n
n
+ 2
n
n
+ 3
n
n
+
ri
n
[a,1884.]
/""
x
(iiL)_J_+
*_+
* .-+ +
-J*/2ri*-"n?
[CLARE, etc., 1882.]
\/2?i-l2 \/4?i-22 \/6rc-32
(iv-)
"
i
n
beinsr K sin2/" + sin2K + sin2fc +... +sin2K" !-, 2 J 2n 2n 2n (.
"
"
"
an
integer.
28. Show
[ST.JOHN'S, 1886.]
that the limit when
n
is increased
of indefinitely
n
'2n
3n
n2
2*
[COLLEGES, 1892.]
29. Show that the limit when
n
is infinite of
i
i
/*"+*.
is
e^a
this result to find the limit of
Apply
-('+
[CLARE, etc.,1886.]
134
INTEGRAL
CALCUL
US.
30.
Find
the
limiting limiting
the
n
value
of
(n\}n/n
is
when
n
is
infinite.
.
31.
Find
sum
the of
value
when
n
infinite
of
the
Tith
part
of
the
quantities
n
'
n+1
+
2
n
+
3
n+n
~
Ti~J
T~J
"V
n
and
show of
that the
it
same
is
to
the
limiting
as
value
3e
:
of
the
?ith
e
root
of base
the of
product
the
quantities
8, where
is
the
Napierian
32.
logarithms.
is
[OXFOKD,
1886.]
If that
na
always
equal
value
to
unity
of the
and
n
is
indefinitely
great,
show
the
limiting
product
[OXFORD.
1883.]
CHAPTER
IX.
/
EECTIFICATION,
113. propose the In the of the
next
ETC.
course
four
chapters we
of obtaining foregoingmethod of the limit of a summation by application to the problems of findingthe process of integration bounded by such lengths of curved lines,the areas of solids of and volumes lines, finding surfaces
to illustrate the
revolution, etc.
114. As idea order
we
Rules
for the
Tracing
curve
of
a
Curve.
some
shall in many cases of the shape of the
to
have the
to form under
rough
in discussion, of
integration, author's refer the student to the we larger may Treatise on the Differential Calculus, Chapter XII. for a full discussion of the rules of procedure. The followingrules, however, are transcribed for
properly assign
,
limits
convenience
will reference, and suffice for present requirements:"
of
in
most
cases
115.
1. A
curve.
I
For
Cartesian will
Equations.
detect
glance
suffice to
symmetry
in
a
136
INTEGRAL
CALCULUS.
(a) If
no
of y occur, the curve is symmetrical powers larly Simiwith respect to the axis of x. for symmetry about the ^/-axis. odd
=
Thus
y2
4"ax is
symmetricalabout
the cc-axis.
occur
(6) If
of both x and y which all the powers be even, the curve is symmetrical about axes, e.g.,the ellipse
both
^ y*_ a2+62~
(c)Again,if on changingthe signsof x and y, the there remains unchanged, equationof the curve is symmetry in opposite quadrants,e.g., the a2,or the cubic x3+y3 3ax. hyperbola xy If the curve be not symmetricalwith regard to tion either axis, consider whether any obvious transforma= =
of coordinates
2. Notice
could make the
curve
it
so.
whether
passes
through the
the coordinate the points where it crosses coordinates present whose axes, or, in fact any points themselves as obviously the equation to the satisfying
origin ; also
curve.
those parallel to the asymptotes; first, axes ones. ; next, the oblique 4. If the curve equate to pass through the origin of lowest degree. These will terms the terms zero givethe tangentor tangentsat the origin. the
5. Find
3. Find
dx'
y^; and
where the the
it vanishes
or
becomes
in-
find where finite; i.e.,
6. If
we can
or tangent is parallel
pendicul per-
to the #-axis.
of the of the other, x, it will be variables, say y, in terms in the solution, found that radicals occur frequently and that the range of admissible values of x which solve
equation for
one
givereal values ofloops upon a
for y is therebylimited. The existence is frequently detected thus. curve
RECTIFICATION,
7. Sometimes reduced to the 116. II. For the
ETC.
137
when simplified
is much equation polarform. Polar Curves.
some
It is advisable
"
to follow
such
routine
as
the
: following 1. If possible, form a table of r and 9 which the satisfy
of
values corresponding for chosen
etc.
curve
values both
of 9, such
as
6
=
0,
"^, "","f
O
JP
Consider
,
o
and positive
2. Examine
values negative
whether
of 9.
initial line. 9 leaves the
r
=
there be This will be so when
symmetry a change
the of signof
about
unaltered, e.g.,in equation
be frequently
the carclioide
a(l
"
cos$).
obvious from the equation of the curve confined that the values of r or 9 are between certain limits. If such exist they should be ascertained, asijm9, it is clear that r must e.g.,if r lie in magnitude between the limits 0 and a, and the
3. It will
=
curve
lie wholly within whether circular.
or
4. Examine
the circle r a. has any the curve
=
asymptotes,
rectilinear
RECTIFICATION. 117. The process of between two curve
a
of the lengthof an arc finding fication. specified pointsis called recti-
the differential coefficient Any formula expressing of s proved in the differential calculus gives rise at in the integral to a formula once by integration calculus for findings. add a list of the most We common. (The references are to the author's Diff. Gale, for Beginners.} 118.
values
In each of the
the limits of integration are variable corresponding to independent
case
the
138
INTEGRAL
CALCULUS.
the two points which is sought.
terminate
the
arc
whose
length
Formula
in
"he Diff. Calc.
Formula
in the Int. Calc.
Reference.
Observations.
P. 98.
For
Cartesian
Equa
tions of form
P. 98.
For Cartesian Equa tions of form
*=/(*/)"
P. 103. For Polar of form
Equation
P. 103.
For Polar of form
Equation
ds
_
dt
M(di)+(Tt
rdr
l(dx\* Idy
P. 100.
For is
case
when
curvi
given as
use
Pp. 103,
105.
For
when
Peda
dr
Equation is given
For
use
ds
P. 148.
when
Tan
Pola gential Equation is given
119. We
Ex.
1.
add illustrative
the
: examples
"
Find
length of
to
one
the
arc
of the
xL=kay parabola
the latus-rectum.
x"%a.
extendingfrom
y="
,
the vertex and
,
extremityof
are
yi="
the limits
#=0
and
Hence
140
Ex. p
=
INTEGRAL
CALCULUS.
4.
Find
the
rsina and Here
r2.
between
length of the the points at
=
arc
of the the
equiangularspiral
radii
vectores
are
which
arc
f 2- -^r
J
=r
VV2-r2sin2a
any
arc
Ex. whose Here
5.
Find
the
length of
=
of the
involute
of
a
circle,
equation isp
s
=
where of the
^
arc
and
\^2are the values respectively.
of
^
at the
beginning
and
end
120. In
Formula
for formula
Closed
Curve.
using the
in the the
case
of
a
closed be
oval, the originbeing within
that the
curve,
it may is
observed
length of
the
the
whole
contour
given by
I pd\fs, for
"
portion
\~dj)~~]
-jy
when disappears
the limits
are
taken.
Ex.
e
Show
that
the
"
perimeter of
of its
an
of ellipse of
a
small
tricity eccen-
exceeds
by
length that
circle
having
the
same
area.
[7, 1889.]
Here where Hence
p2 ^
is the
=
+ 62sin2^ a2(l e2sin2^ ), c^cos2^
=
-
angle
=
-
which
p
makes
with
the
major
axis.
p
.A i^sin2^ I a( -e*sui*\ls.
-
\
2i
o
/
Hence
I \-^-\\ ,=4a{|-lA f}(very approximately)
RECTIFICATION,
The radius of (r)
a
ETC.
141
circle of the
same
area
is
given by
vZ^ab^a^l-erf,
/
I
o
"
"
\
and
its circumference
=
1 27ra( \
"
-e2
4
e4
...
).
/
32
.
*.
Circumf.
o
' =
"
ellipsecircumf
-
circle
.
=
( \lo
"
-
\irae*
=
_
""
.
2ira
3'2
/
t"4
4
of circle], far as [circ.
as
terms
e4. involving
64
EXAMPLES.
1. Find
=
the length of the arc by integration where the points between a2,intercepted
of the
x"a
cos
circle and a
2. Show
that
in
the
catenary y
=
c
cosh
-
the
length of
arc
from
the vertex
(where #=0)
s
=
to any
c
point is given by
x
-.
smh
.
i
c
v
3. In
the
that where
the
evolute of length of the the
it meets
viz., 4(# 2a)3 27a#2,show parabola, from its curve cusp (# 2a) to the point is 2a(3v3 1). parabola
a
-
=
=
"
4. Show
that the
lengthof
the
arc
of the
cycloid,
.r=a(0+ sin 0) ^ a(l-cos"9)J y
=
between
5. Show
the
pointsfor
which
0=0
and
0=2^,
is
s
=
that in the
for epicycloid
which
y=(a
+
0 6)sin
-
b sin
=
a
26
5
measured beiijg
n
from show
the
point at
222
which
0=7rb/a.
and that x*. if
5
When measured
"=--,
from
a
that
4r+y*"a*j
on
be
cusp which
lies
the
s3 oc y-axis,
142
6. Show
INTEGRAL
CALCULUS.
may
that in the ellipse # be expressedas
=
a
cos
$, j/
=
6sin^, the perimeter
7. Find
the
r (i.)
length of
=
any
arc
of the
curves
acos0. aem0.
r (iii.)
=
a6.
asin2-.
2t
r (ii.)
=
r (iv.)
=
8.
Apply
the formula
s=_"
-f cos
+
to \pdty
the rectify
cardioide
whose
equationis r=a(l
radii vectores
0).
of the
curve
[TRINITY,1888.}
9. Two
OP, OQ
equallyinclined to arc length of the intercepted of the measure anglePOQ.
are
drawn
the
initial line
a
; prove
that
is act, where
is the
the circular
[ASPARAGUS, Educ.
an arc
Times.']
can
10. Show
that the
lengthof
of the when
curve
yn=xm+n
or
"
be
found
in finite terms
in the
cases
"
+
-
is
an
integer.
11.
*m
*m
*
Find
curve
the
of the
length of the arc between (c2 a2)p2=c2(r2 a2).
"
"
two
consecutive
cusps
12. Find
the whole
lengthof
the
loopof
the
curve
3ay2=x(x-a)2i.
13. Show
[OXFORD, 1889.]
=
a? that the lengthof the arc of the hyperbola xy and x=c is equal to the arc of the between the limits x=b the limits aV2 between curve r=b, r=c. ""2(a4+r4)
=
[OXFORD, 1888.]
14. Show
that
in the
arc
parabola
and "=1+0080,-^ =:-__^_
T
'
d"Y
sin^w*
hence
show
that the
between intercepted
extremity of
the latus rectum
and the_yertex is a{\/2-flog(l +\/2)}.
the
[I.C. S., 1882.]
RECTIFICATION,
121.
ETC.
143
Length
of the
Arc
of
an
E volute.
It has been shown Gale, for Beg.,Art. (Diff. that the difference between the radii of curvature
157)
at
Fig. 12.
two
pointsof a arc corresponding
curve
is
equal to
the
lengthof
the
of the evolute ;
if ah be the arc of the evolute of i.e., of the original then (Fig. curve, 12)
.e.
the
portionAH
(at H),
(atA)
"
/"
144
INTEGRAL
CALCULUS.
regardedas a rigid curve, and a from it, be unwound then the string beingkept tight, pointsof the unwinding stringdescribe a system of of which is the original AH. curves one curve parallel
e
and
if the
volute be
Ex.
Find
the
lengthof
the evolute
of the
Let a, a', be the centres of curvature /3, /3' the extremities of the axes, viz., A, A', B, B' of the evolute to the arc arc a/3 corresponds and we have (Fig. 13)
arc
ellipse. to corresponding
respectively.The
AB of the curve,
a/2 /o(at 5)-p(at A)
=
=
~-
"
rad. [for Thus the
of
curv.
of
ellipse
=
^.
Ex. 3, p. 153,Diff. Beg."]. Calc.for the evolute
lengthof
the entire
perimeter of
EXAMPLE.
in the above manner for the parabolay2 kax that the within the parabola lengthof the part of the evolute intercepted Show
=
is4a(3\/3-l).
122. Intrinsic The
Equation.
s,
relation between measured
the
given curve,
from
lengthof the arc of a a given fixed point on
Fig. 14.
the curve, and the extremities of the of the
curve.
anglebetween
arc
the
is called the
tangentsat the Intrinsic Equation
RECTIFICATION,
123. To
ETC.
145
obtain
the Intrinsic
Equation
from
the
Cartesian.
be given as y=f(x). of the curve Let the equation and Supposethe #-axis to be a tangent at the origin, from the origin. the lengthof the arc to be measured Then also
tan
-^=/("),
(1)
s=\ *Jl+ [f'(x)~]2dx
(2)
from (2), If s be determined and x by integration eliminated between this result and equation (1), the relation between s and ^ will be obtained. required
.
Ex.
1.
Intrinsic
equationof
a
circle.
If
be i/r
tangent
the at the
angle between
point P,
and
a
the initial tangent at A and the have the radius of the circle, we
and
therefore
2.
s"a^r.
of the
c
Ex.
the case equationis s
=
In
catenary y
+
c
=
ccosh-,
c
the
trinsic in-
tan
^.
=
For
tan^
as
"
^
dx
and
T
=
\/ 1
"
/-,
-r
smh2-
"
i
""x
=
dx
P. I. c,
c
K
146
INTEGRAL
CALCUL
US.
and
therefore
s
=
c
sinh
-,
c
of the constant whence together,
being chosen integration
s
=
so
that
x
and
s
vanish
c
tan
\js,
124.
To
obtain
the Intrinsic
Equation
from
the
Polar.
Fig. 16.
/
",
to the the initial line parallel pointfrom which the arc is measured.
Take
tangent at
Then
the
with the
usual notation
T
we
have the curve,
.......
the equationto =/(0), 0+0, ^
=
.......................................
(1) (2)
from and 0, "f" (4), by integration of equations(2) and eliminated (3),the by means will be found. relation between s and \fs required If
s
be
found
Ex.
Find
the intrinsic
equationof the r=a(l -cos 0).
cardioide
Here and
i/r
a sin
0
2
148
INTEGRAL
CALCULUS,
125. When
the
Equation
of the Curve
is
given
as
we
have
tan
dy ^
=
d)'(t)
=
,-
-^ ax
j (t)
^_-^
.......
.....
........
(1)
N
By
means
of
equation(2)s
may
be
found
by
tegrati in-
of t. the result and If then, between shall obtain the eliminated, we
in terms
equation(1) t be required relation
between
Ex.
s
and
^.
In the
cycloid y"a(\ t\
*
= ,
-cos
we
have
tan
^
*mt
=
tan 2
1+costf
Also
^ dt
5=4a
=
+ cos02+ ax/(l
sin2*
=
2a
cos
-,
2
whence Hence
sin
4ot sin
-
if s he measured
from
the
where origin
Z=0.
5
=
is the equation T/T required.
126. Let
Intrinsic
Equation
of the Evolute.
be the equationof the given curve. s=f(\f/) of the arc of the evolute measured Let s' be the length fixed pointA to any other pointQ. Let from some the original 0 and P be the points sponding correon curve to the points A, Q on the evolute;p0, p the P: at 0 and radii of curvature \j/the angle the and ^ the tangent QP makes with OA produced, with the tangent at 0. the tangentPT makes angle
RECTIFICATION,
Then
and "*//-^r,
=
ETC.
149
ds
or
O
T
*
Fig. 18.
Equation of an Involute. if the curve the same With AQ be givenby figure, have the equation we s'=f(\}/),
127. Intrinsic
and whence
Ex. Hence The intrinsic
\Is
"
\//,
=
\
equation
of the
catenary
is
is s=ctsm\Ir
(Art. 123).
the intrinsic of equation its evolute
and
p0=
=
radius
c
of curvature
= =
at the vertex
p
y
-
c
and sec2i/r
s
=
T/r=0
,
' . .
the evolute
The
1 ),or c(sec2-v//intrinsic equation of an involute
is
s
=
c
tan2^.
is
s
=
I(ctan "fy + A)d^r
c
=
+ A^r + logsec T/T
constant
;
we
and
if s be
so
measured
s
=
that 5=0
c
T log(sec
when
^=0,
have
150
INTEGRAL
CALCULUS.
128.
Length of Arc
of Pedal
Curve.
the
originupon the tangentto any curve, and ^ the angle it makes with the initial line, we regardp, % as the current may polarcoordinates of a pointon the pedalcurve. culated Hence the lengthof the pedal curve be calmay by the formula
If p be the
from perpendicular
Ex. of the
Apply the above method to find the lengthof pedalof a circle with regard to a point on the a cardioide). (i.e.
any arc circumference
Fig.
19.
Here, if 2a
be the
diameter, we
p
=
have
=
from
the
figure
OP
*
cos
2ctcos2*.
Hence
arc
of
pedal
=
=
/2 A/a2cos4+ a2 sin2-
cos
-a
J
*
2
2
2 + C.
/2a
j
cos
*dx
=
4a sin
The
limits for the upper half of the Hence the whole perimeter of the
curve
are
x
=
0 and
X
=
TT.
pedal
=8a.
1 2[4asin-Jo
L-
2
RECTIFICATION,
ETC.
151
EXAMPLES.
1-. Find 2. Find
the the
length of
any
arc
of the
curve
fu\a x)=aP.
"
[a,1888.]
lengthof the
y
=
givenby completecycloid
a "a cos
0. 1
3. Find
from
for which the lengthof the arc measured the originvaries as the square root of the ordinate. the
curve
4. Show
that the intrinsic
s
=
equation of
a
the
parabolais
a
tan
^
sec
^r+
+ sec ^). i/r log(tan
5.
Interpretthe expressions
the wherein closed given
line
curve.
are integrals
taken
rou
id the
of perimeter
a
[ST.JOHN'S, 1890.]
axis of
an
Jw
6. The
major
is ellipse
1 foot
in
length,and
its feet
is 1/10. ^/^eccentricity
Prove
its circumference
to be 3*1337
nearly.
7. Show cardioide
4r=3asec 8. Find
[TRINITY,1883.]
that
r
=
the
a(l +
0 remote
the
length of the arc of that part of the cos 0), which lies on the side of the line is equal to 4a. [OXFORD, 1888.] from the pole,
an arc
lengthof
of the cissoid
r_asin26"
cos
ff
curve
9. Find
the
lengthof
any
arc
of the
10. Show
that the intrinsic
=
3a3/2=2^ is 9s
11. In
a
equation of 4a(sec3Vr1).
-
the semicubical
bola para-
certain
curve
show
that
5=ee\/2+
a
152
INTEGRAL
CALCULUS.
12.
Show
that
the
length
of
an
arc
of the
curve
is
given by
13.
s
=/("9) +/"(#)
curve
+
C. the intrinsic
Show is
s
=
that
a
in the
y
=
alogseca
tion equa-
gd~l\ff.
that the
14.
Show the
length
of the
arc
of the
curve
y=logcoth-
between
is log s!n x^. yj),(#2, 3/2) points (xl9
sinn
X-^ the
15.
Trace
the
curve
y2
which
=
g"
(a
"
#)2,and
find the
length
of that
od/
part of the evolute
16.
=
corresponds to
of
an arc
loop.
1881
[ST. JOHN'S,
Find the
and
1891.]
(p
equiangular spiral pole. Show that the arcs of an from equiangular spiral measured the pole to the different with another points of its intersection pole but a different angle equiangular spiralhaving the same will form in series a [TRINITY, 1884.] geometrical progression.
an
length
of
rsma)
measured
from
the
17. has
Show
that
the
curve
whose
s
=
pedal equation
a"-.
Zi
is
p2=r2
"
a?
for its intrinsic Show
to
equation
whole
18.
that that of
the
an
length
of
the
is
equal
whose ellipse minimum
semi-axes
limagon r=acos are equal
of the of
a
in
length
to the 19.
curve
maximum Prove that
and the
radii of the
vectores
limacon.
length
nth
pedal
loop of
the
rm=amsinmO
is
,-m
mn-m+1
a(mn+I)
o
(smmO)
of
m
dO.
^
of the
curve
1883
j
20.
Show
that
the
length
a
loop
[ST. JOHN'S, 1881.]
CHAPTER
X.
QUADRATURE,
129. The
ETC.
Areas.
Cartesians.
bounded process of findingthe area is termed quadrature. portionof a curve It has bounded ordinates considered for the been
by
the
any
area
already shown
any curved and
x
=
in Art. 2 that
by [x
as
line
[y
sum
=
"f"(x)], pair of any
axis
an
=
a
b] and
the
the limit of the is
of
of x9 may be infinite number
of inscribed
area
and rectangles;
that
the
expression
1 ydx
or
0 (x)dx.
In two
the
same
way
the
=
area
bounded
=
given abscissae [y
c, y
d]
and
by any curve, the y-axisis
fxdy.
by two Again,if the area desired be bounded given curves [y "p(%)and 2/ \^(^)]and two given ordinates it will be clear by similar \x a and x 6], reasoningthat this area may be also considered as the limit of the sum of a series of rectangles constructed
130.
=
=
"
=
154
INTEGRAL
CALCULUS.
as area
indicated will
figure. The be accordingly
dx
or
-
in the
expressionfor the
Li% PQ
x=a
fj"(0) \fs(x)]dx.
J
Fig. 20.
Ex.
1.
Find
x=c,
area=
the x"d
area
bounded the
"- + by the ellipse 2
^2
-
=
1, the
b2
ordinates
and
Here
J
f ^Sr
a
2a
For the
we quadrant of the ellipse above expression becomes a
must
put d
=a
and
c=0
and
"
a2
. .
?
2
or
^"
.
2a
4
givingirab
Ex.
curves
for the the
"
area
of the whole above
=
ellipse.
included
between
cut to
x"a.
2.
Find
=
area
the #-axis
the
at
y2 ax. and parabolatouch at the origin So the limits from of integration #=0 are (a,a). The area is therefore sought
%ax
y2
x2 and the
The
circle and
again
fa
?
-
x2
~
156
For the
"
INTEGRAL
CALCULUS.
portionbetween
a
the
as
curve
and
the
asymptote the
limits are For the
to
0, and double
therefore
before.
loopwe
have
a+x
for the
between portion
the
curve
and
the
/v.
_
asymptote,
x\l /O
In
dx.
" a+x
Fig.22.
To
integrateIxJa~xdx, put
J
*
a ~p x
x=a
cos
0
and
dx"
"a
sin 0 c?$.
Then
J
ftfJEfcfcl\ Va+x
-
J
1-cos2^
r
=
a2/
and
area
of
--}" I)
QUADRATURE,
ETC.
157
Again,
rxJ?E*dx=
J_
"a+#
-
a
cos
J
"^gain fl^1 l-cos2#
"
Bd9
[The meaning
+
of the
negativesignis
in
_
this
:
"
In
are
the choosing
sign before
the
radical
y=#/v/^"
*
we
tracing the
a+x
the curve below the #-axis on the left of the origin and above the axis on the rightof the origin. Hence y being be is it referred to between limits the expected to, negative that we should obtain a negative value for the expression
portionof
Thus
the whole
area
is required
in this example that the greatest also be observed assumed that infinite one. In Art. 2 it was ordinate is an for the the result area finite. Is then was every ordinate ? t rue and the the bounded curve asymptote rigorously by between limits let us integrate To examine this more closely small positive so as e is some a + e and quantity, 0, where have to exclude we the infinite ordinate at the point x" "a,
[It must
"
as
before
J
where
so
A/fEfdfc. [""c
"
*a+x
that 8 is
a
small angle. positive
This
is integral
4
2
4
which
close approaches indefinitely
to the former
result
when
8 is made
to diminish
without
limit.]
158
INTEGRAL
CALCULUS.
EXAMPLES.
1. Obtain rectum.
the
area
bounded
by
a
parabola and
the
its latus
bounded the areas by the curve, ordinates in the following cases : specified
"
2. Obtain
and ^7-axis,
the
(a) #=ccosh-,
to x=h.
x= x=a a
to x=b. to x"b.
3. Obtain 4. Find
the the
=
area areas
bounded
the
of X2la2+y2/b2l is divided
5. Find
by the curves y2=4ax, #2=4ay. portionsinto which the ellipse
the
curve
by
the whole
area
the line y=c. included between
"
X2y2=a2(y2x2)
and its
asymptotes.
the
area
6. Find
between
the
curve
y2(a+x)=(a
"
xf
and
its
asymptote.
7. Find the
area
of the
loopof
the
curve
y*x+ (x + af(x + 2a)
=
0.
131. When
Sectorial the
area
Areas.
Folars.
is bounded by a curve r=f(6) and two radii vectores drawn from the origin divide the area into elementary in givendirections, we small angle89, as shown in the sectors with the same figure.Let the area to be found be bounded by the arc
to be found
PQ and the radii vectores OP, OQ. Draw radii vectores OP19OP2, OPn-i at equal angularintervals. Then by drawingwith centre 0 the successive circular arcs it may be at once that the seen PN, P1NV P2^2,etc., of the circular sectors OPN, OP^N^ limit of the sum is the area required. For the remaining OP2N%, etc., elements PNPV P^N^P^ P2^2P3,etc., may be made to the rotate about 0 so as to occupy new on positions
...
QUADRATURE,
ETC.
159
greatestsector say OPn-iQ as indicated in the figure. Their sum is plainly less than this sector ; and in the limit when the angle of the sector is indefinitely
diminished its
area
also diminishes
without finite.
limit provided
the radius
vector
OQ
remains
O
Fig. 23.
The
area
of
a
circular sector is
of angleof sector. X circular meas. J(radius)2 the summation Thus the area required l?L"Zr2S(), being conducted for such values of 9 as lie between Ox being xOP and 0 6 xOPn-i, i.e., xOQ in the limit,
=
=
=
the initial line. In and the notation
=
of the will be
or
xOQ
this /3,
calculus integral expressedas
if xOP
=
a,
Ex. and Here
1.
Obtain
the
area
of the semicircle the
bounded
by
r
=
acos
0
the initial line. the radius vector
to
sweeps
area
over
angularinterval
from
0=0
0"
-.
Hence
the
is
2
i.e., ^radius)*.
160
Ex. 2. Obtain
INTEGRAL
CALCULUS.
the
area
of
a
loopof
the
curve
r"a
sin 3ft
This curve will be found to consist of three equal loopsas indicated in the figure (Fig. 24). The proper limits for making the integration extend over the first loop are 0=0 and 6 of 0 for which r vanishes.
=
^
-,
for these
are
two
successive values
.-. area
of
loop 1 fWn2
=
30 dO
=
~
f\l
-cos
60)d9
4
3~~ 12'
2
The
total
area
of the three
loopsis
therefore!^.
Fig. 24. EXAMPLES.
Find
1. r2
=
the
areas
bounded
by
3. One
r=
r
=
"2cos20+ 62sin2ft
2. One 5. The
9=13 and
6.
loopof 4. One loopof loopof r=asin2ft bounded by the portionof r=ae^coia 0=/3 + y (y being less than 2?r).
sector sector sector
a a
sin 4ft shift ft
radii vectores
Any
of of of
7.
8.
Any
Any
7^0=^ ((9=a to 6*=^). r0""a (0=a to ^=)8).
r(9
=
9. The 10. If
cardioide
s
r
=
(9=a to a(l cos 0).
a
"
0
=
fi).
between
2
be the
lengthof
A
=
the
curve
r="tanhbetween the
the
originand
show
$
=
27r,and A
the
"
area
same
points,
1888.
that
a(s air).
[OXFOKD,
]
QUADRATURE,
132. Area Let P
on
a
ETC.
161
of
a
Closed
Curve.
(x,y) be the Cartesian coordinates of any point closed curve ; (x+ Sx,y + Sy)those of an adjacent
Let
Q. point
(r+ (r, 9),
Sr , 6 +
$0)be the corresponding
shall suppose that in Also we polar coordinates. from P to Q along the along the curve travelling infinitesimal arc PQ the direction of rotation of the OP that the is counter-clockwise radius vector (i.e.
Fig.25.
hand to a person the left Then the element direction).
area
is
on
in this travelling
ir2(S$ Hence
curve
=
AOPQ
f
=
$(xSy ySx).
"
another is
for expression
the
area
of
a
closed
Wxdy-ydx),
beingsuch round the completely
the limits 133. If
we
that the
curve.
point(x,y) travels
that
once
put
y
=
i
so
^M^
=
we (fo)
may
write
the above
where as ^\xzdv, expression
terms
x
is
the limits of integrati that the current point (x,y) travels so chosen As v is really once completelyround the curve. 6 is a rightanglecare tan 6 and becomes infinitewhen be taken not to integrate must throughthe value oo
to be in expressed
of
v
and
.
E. i. c.
L
162
Ex.
INTEGRAL
CALCULUS.
Find
by
this method
the
area
of the
ellipse
#2/a2+.y2/"2=l. Putting y
=
vx,
we
have
and
=
*
JV
between chosen properly
f"^L= f*
L2
Now,
limits. in the first quadrant v varies from
area
0 to
oo
Hence
.
of of
quadrant =?"
"
-,
and
therefore
area
ellipse
=irab.
134. If the
current
originlie
without
we
the
curve,
as
the
elements such
as
triangular of space such as OP1QV including portions lie outside which OP2Q2 shown in the figure
point P
travels round
obtain
Fig.26.
the
curve.
removed travels element and S6
These portions however are ultimately from the whole integral when the point P the element over P2Q2, for the triangular 9 is decreasing
as OP2Q2 is reckoned negatively is negative.
135. If however
the
curve
cross
the expression itself,
no perimeter,
^ I(xdy
"
ydx), taken
round
sum
the whole
areas
the longerrepresents
of the
of the several
164
INTEGRAL
CALCULUS.
the curve is when adapted to the cases specially defined by other systems of coordinates. Ss of a plane curve, and OF If PQ be an element the chord the pole on from the perpendicular PQ,
Fig. 28.
|OF.PQ,and any sectorial area summation the along the whole being conducted of the Integral culus CalIn the notation bounding arc.
AOPQ
=
=
this is
[Thismay
be at
once
deduced
from
| rW,
ds
thus
:"
(V2d0ir^ds sin 0 \r
= =
is (where "f"
the
angle between
the
tangent
and
the
radius vector)
137.
Tangential-PolarForm.
ds
d*p
Again,
we
since
P
=
=
5^
P +
have
area
=
ds \ \p
.
.
=
J
QUADRATURE,
a
ETC.
165
formula
suitable for
use
when
the
Tangential-Polar
equationis given.
138. Closed When
some
Curve. is closed this admits expression of
the
curve
simplification.
For and
term in
round integrating Hence disappears.
area
=
the whole when the
the perimeter
curve
first
we
is dosed
have
^
the equation of Ex. C ale.forBeginners, By Ex. 23, p. 113, Diff. the one-cusped be the as (i.e., cardioide) expressed epicycloid may p
=
Fig. 29.
Hence its whole
a2cos2^ \d^ taken area=-^/f 9a2sin2;'
"
tween be-
limits
i/r 0
=
and
^="
becomes
IT
and
doubled.
Putting-^ 3$,this
= =
3a2
f (9sin2^
^o
-
co**0)dO
=
67ra2.
166
INTEGRAL
CALCULUS.
139.
Pedal for
Equation.
curves
Again,
we
given by
dr
their
pedal equations,
have A
=
ip
ds
=
i p
=
}p
sec
0 dr
sin
a,
=
i
-
Ex. Hence
In the any
equiangularspiral p=r
sectorial
/"2
=
area
y2si
rcosa
f
J
/
140. Area of curvature In this
case
included and the
we
between evolute.
as our
a
curve,
two
radii
take
element
two
arc
of
area
the
contained elementary triangle of curvature and
by
contiguousradii
ds of the
curve.
the infinitesimal
To
first order
t
infinitesimals
this
is
and |/o2"S\^,
or
the
area
=
p, between
a
.e.
p\
Ex.
to the
1.
The
area
and its involute, circle,
a
tangent
circle is
(Fig.31)
QUADRATURE,
ETC.
167
the tractrix and its asymptote is between Ex. 2. The area found in a similar manner. such that the portionof its tangent The tractrix is a curve and the ^7-axis is of constant between the point of contact
length c.
Fig.31.
Taking two adjacenttangents and elemental triangle 32) (Fig.
the axis of
x
as
forming an
o
T
r
Fig. 32.
EXAMPLES.
1. Find
the
area
of the
two-cuspedepicycloid
[Limits \jf 0
=
to
"^=7rfor by
=
one
quadrant.] pedalequation
2. Obtain
the
same
result
means
of its
7.2 ^2 +
r=a [Limits
1^2.
one
to
r
=
2a for
quadrant.]
168
3. Find
INTEGRAL
CALCULUS.
the
area
between
at
the radius
curvature. 4. Find
of curvature
catenary s the vertex, and
the
the
=
c
tan
^,
its
evolute,
of
any
other
radius
the
any
area
between
between the
s epicycloid
=
AsmB^s,
its
and evolute,
5. Find
two
radii of curvature.
the
area
s"Ae^y spiral equiangular
its
and evolute,
any
two
radii of curvature.
AREAS 141. Area If of Pedal
OF
PEDALS.
Curve.
be the tangential-polar _p=/(Vr) equation (Diff. Gale, for Beginners,Art. 130) of a given curve, S\fs will be the angle between the perpendiculars two on and the area of the pedal may be contiguous tangents,
as expressed
(compare Art. J|p2c^/r
131).
Fig.33.
Ex. Find the
area
of the
pedal of
a
circle with
regardto
a
(thecardioide). pointon the circumference if OF be the perpendicular Here the tangent at P, and on OA the diameter obvious that OP ( 2a), it is geometrically
=
bisects the the
Hence, calling angle AOY. YOA"^, of the circle tangential polarequation
we
have
for
Hence
=
^/
QUADRATURE,
ETC.
169
where the limits are to be taken as 0 and TT, and the result to be doubled so as to include the lower portionof the pedal. Thus
*cos*fe^ 4a2. ^l=4aaf 2 J
=
2
f
J
o
4222
o
Fig. 34.
142. Locus Let 0 be of
a
of
Origins of
Pedals
of
given
Area.
tangentto
a
be the polarcoordinates point. Let pt \js the foot of a perpendicular OF upon any givencurve.
fixed
o
0
Fig.35.
Let P
be any from
other P
fixed the
upon
the perpoint, J"F1(=^1) pendicul tangent. Then the areas
170 of the
INTEGRAL
CALCULUS.
with pedals
0 and
P
as are origins respectively
and j[ftL2"% ijV2cfyr taken
areas
between
A
certain
and Al coordinates of P
Call limits. Let r, 6 be the respectively definite with
these
polar
their
regard
"
to
0, and
x9
y
Cartesian
Pi and p is
2 Al
=
"
equivalents.Then T cos($" -t/r) P =p
~~
x
cos
ifs
"
y sin
i/r,
a
known
=
function
"
of
\fs Hence
"
\(p \p^d\fr
"
x
cos
i/r
"
y sin
\l/fd\^
=
Vp^d"^2x \pcos ^ d\/r 2^/ d\fs |psin \fs
1 cos + x* I + 2a32/ cos2i/r c?i/r
+
si \[s
1 si 2/-2
cos
Now
2
Jp
I sin \/r 2 Ip d\[s, d\fs, \/r
between
such limits that the whole pedal is described Call them will be definite constants.
-20,
and
we
-2/,
a,
2A, 6,
ax2 +
thus
obtain
=
2Al
"2A +
2gx + 2fy+
If then P move its locus must
in such a manner be a conic section. of Conic, result in
2hxy + by2. that Al is constant,
143. Character It is
a
known
that inequalities
Hence
it will be obvious
that if p, q, r,
...
stand
,
for
172
INTEGRAL
CALCULUS,
will
thereby be
removed.
Thus
"2 is
a
point such
that the
both and \psiu\^d\[^ \pcos\fsd\fs integrals if II be the for any
=
vanish, and
is "2
we
area
of the
pedal whose
pole
have
other 211 + ax2 + The
area
in the
2A1 case. general
2hxy + by*
of this conic is
*
"
Art. 171). Conic Sections, (Smith's -d.1 IH
=
Thus
I?
"
A
TT
*/ab
"
h2
,
""
,
-"
s
(area or conic).
closed oval the equation
\
^7T
For
the
case particular
of any
of the conic becomes
whence where
r
J.1
is the radius of the circle values of
on
which
P
lies for "2.
constant
Av
i.e.the distance of P
from
146. Position In any oval
at plainly
of the which
Point
has
a
"2 for Centric
centre
Oval.
is
as
the
point "2
is taken
that centre, for when
the centre
the integrals and \psui\fsd\ \pcos\fsd\fs origin, both vanish when
is performedfor the integration elements of the integration (opposite
the
complete oval cancelling),
147centre
Ex. any
1.
Find
the
area
regard to
Here and Hence
point within (a limagon). A
n
=
the
of the pedal of a circle with circle at a distance c from the
n+^,
7ra2.
.
=
Ai=ica*+"
QUADRATURE,
Ex.
2.
ETC.
173
with regard ellipse
of the pedal of an to any point at distance c from the centre. of the pedalwith In this case II is the area Find the
area
-
regardto
the centre
2
/* Vcos2"9 + b%m*0)dO
=
+ (a2
"2)|.
Hence
Ex. 3.
^1=|(
The
area
of the
an
taken
c
with
respectto
pedalof the cardioide r=a(l "cos 0) internal pointon the axis at a distance
from
the poleis
|(5"s-2"c+2c'). [MATH.TBIpos"187a]
Let 0 the two
pole,P the given internal point; p OF2 and PTl on any tangent perpendiculars
be the and
cos
and pl from 0
and P
pl "p
the angle Y$P respectively ; ""
"
OP"c
; then
c cos
"",and
^Al=2A
"
2clp
+ / cfc" ""
Fig. 36.
Now in order that p may limits sweep
=
out
the whole
pedalwe
Now
must
between integrate
the cardioide
""
0 and
""
=
-^and
double.
in
36) (Fig.
p=
OQ
sin
Y2QO
=
OQ
sin^xOQ.
[Dif.Calc., p. 190.]
174
INTEGRAL
CALCULUS.
For
r2"0
=
itf0"|
=
Hence
|-{*-("-W-|
|-*=f,
/3
or
-
J-J-J,
-
so
"
-.
23
,
A*cos
,
Hence
/p
"j" d"j" 2
=
/
'
2a cos3 2
cos
d(f" "/"
3,so?2
=
4a
x
3
/ cos%
o
fl
cos
=
12#
/
rf [4cos%
-
3
cos")"iz
6422
42
2
Also
^^^ fc2cos2d"cta=3.2c2i
J
Sir
222
Finally
24
=
2
J
Tcos^ "fe, 4a*"*^*J"
=
24a2
6
642
mi
3
1
^
2
?rac
'
A Al
_?ra
--
8~
T"
148. When
Origin for
f2 is taken
2A
l
=
Pedal
as
of Minimum
Area.
that
it appears origin,
cos
211 +
(05 J
^
+ y sin
\Jsfd\ls.
Hence
as
the term
is necessarily \(x + ysm\}s)"2d\fs cos\fs
it is positive,
clear that
Al
can
never
be
less than
II.
QUADRATURE,
"2 is therefore the
a
ETC.
175
the
for origin
minimum
which
area.
corresponding
pedal curve
149. The
has
Pedal formula is
area
of
an
Evolute
area
of
a
Closed
Oval.
for the
of any
closed oval
proved
in Art. 138
of
Hence
J
) jp2*/' Jjft
=
oval +
plainlyexpresses that the area of any pedal of the oval itself is equal to the area of an oval curve togetherwith the area of the pedalof the evolute (for
which
-ry This is the
radius
vector
of the
pedalof
the
evolute).
also admits
of
proof. elementarygeometrical
pedalof
the evolute of
an
of the Find the area Ex. with regardto the centre. The
area
ellipse
above
article shows evolute
=
that
area
of
pedalof
of
pedal of ellipse area
-
of
ellipse
-
-(a2+ b2)
irab
=
?(a b)2.
-
176 150. Area
INTEGRAL
CALCULUS.
bounded
by
a
Curve, its Pedal, and
a
pair
Let
of
Tangents.
two
contiguouspoints on a given 7, F' the corresponding pointsof the pedal of curve, since (with the usual notation) Then any origin0.
bounded PF=-vjrthe elementarytriangle
P, Q be
by
YY'
two
tangentsPY, QY' contiguous
and
the chord
is to
the firstorder of infinitesimals
Fig.38.
Hence
curves
the and be
area a
bounded of any portion by the two curve pair of tangentsto the original
may
as expressed
and
area
is the of the
same
as
the
portion of the corresponding
evolute.
and
pedal of the
151. Let
Corresponding Points f(x, 2/)
=
Areas.
Its
area
0 be
any
closed
curve.
(A^)
QUADRATURE,
is the
ETC.
177
by expressed
\ydxtaken line-integral
round
the complete contour. If the coordinates connected relations the curve
x
of the current point (x,y) be with those of a second point(" rj) by the mg, y nrj, this second pointwill trace out is expressed /(w" nrj) 0 whose area (J.9)
=
= =
by
the
I line-integral rjdg taken
we
round
its contour.
And
have
l=
dx \y
=
\ nrjm
=
whence
y) f(xt
=
it appears that the area of any 0 is mn times that of f(inx, ny)
1.
closed
=
curve
0.
152. Ex.
Apply
this method
to find the
1
area
of the
ellipse
**
+
,"*
~
=, a r
b
r
the
point", T\ traces corresponding
out the circle
and
area
of
ellipse
=~
x area
of circle
r
Ex. Let
2.
Find
the
area
of the
mx
=
curve
(mV
=
+
n^f)2
=
a
V
+
Wif-
^
is
ny
ij,
then the
curve corresponding
or
in
polars pedal of
an
r2
=
^- cos2 0 +
m2
sin20, ri2
"--,
the central
E. i, c.
symmetrical about ellipse,
M
both
axes.
178
Hence the
area
INTEGRAL
CALCULUS.
of the first curve
=
"
x
area
of second
mn
EXAMPLES,
1. Find
the
area
of the
loopof
of the
=
the
curve
ay"L=x\a-x).
2. Find
[I.C. S.,1882.] [I.C. S., 1881.]
prove
a.
the whole
area
curve
3. Trace
the
curve
a?y2 a2x2-x*. y\ and a2#2=;?/3(2ct
"
that its
and
area
is
equalto
that of the circle whose
radius is
[I.C. S., 1887
the curve cfiy*xb(Za x\ and prove radius is a as 5 to 4. to that of the circle whose
4. Trace
=
"
1890.]
area
that its
is
5. Find
the whole
area
of the
curve
[CLARE, etc., 1892.]
6.
By
means
of the
/y integral by
the the
dx
taken
round
the
contour
of the
formed triangle
lines intersecting
show
that
they enclose
area
PKIZE, 1876.] [Sir.
7. Find
8. If
the
area
between
y2
=
and
a
"
its asymptote. with the axis of #,
x
ty be
the
area
angle the tangent
of y
an
makes
show
that the
oval ds "fy
curve
is q:
/r
the
cos
or
/x
the
sin
^rds,
integration being taken
all round
perimeter.
180
INTEGRAL
CALCULUS.
line to a point P on the curve if A be the ; and by the curve, the initial line,and the radius vector 9,42
21.
=
area
bounded
to
P,
then
2rf.
Find
the
area
of the closed
3a sin
_
portionof
0
cos
the Folium
9
~sin^6"TcoW
In what ratio does the line x+y
=
[I.C. S., 1881]
Za
divide
the
area
of the
loop?
22.
[OXFORD, 1889.]
Find the
area
of the and
curve
r=aOebe
enclosed
between of the
curve.
two
given radii vectores
23.
two
successive
branches
[TRINITY,1881.]
Find the
area
of the
loop of
the
curve
r
=
a0cos 0 between
24. Show
that
the
area
of
a
loop of
cases
the
n
curve
r
=
acosn0
is
^"
4?i
-,
and
state
the total of
area
in the the
odd, n
r
=
even.
25.
Find
the
area
a
loop of
curve
a
cos
3$ + b sin 3$.
[I.C. S., 1890.]
that the 26. Show the curve r=acos5$ circle. 27. Prove that the
area area
contained
between
the circle of the
r=a
and of the
is
equal to three-fourths
of the sin 0
curve cos
area
[OXFORD, 1888.]
2ac r2(2c2cos2"9
-
9+
a2sin2#) aV
=
is
equal to
28.
irac.
[I.C. S., 1879.]
whole
area
Find
=
the
acos
of
the b
"
curve
representedby
the
two
the
equation r
29.
curve
0+
area
b,assuming
included
a.
Find
r
=
the
between
loops of
[OXFORD,
the
a(2cos 0 + ^3).
the
area
1889.
]
30.
Find
between
the
curve
r=a(sec $+cos 0)
of the
and
its
asymptote.
Prove r2 that
=
31.
the
area
of
one
loop
pedal
of
the
lemniscate Find
a2cos2$
with
respect to the
poleis a2.
[OXFORD, 1885.]
32.
the
area
of the
loop of
of the
the
=
curve
(x'\-y)(x^+y2) ^axy.
33.
[OXFORD,
curve
1890.]
Prove
that the
area
loop of
the
QUADRATURE,
34.
ETC.
181
Find
the whole
area
contained
between
the
curve
and
its
asymptotes.
that the the the
area
[OXFORD, 1888.]
of the the
35.
Show
(*L^ a2+b2-r2 ellipse p2
=
in-
eluded
vector
between from
curve,
semi-major axis,and
a
radius the
r
centre, is
"
tan"1^/^^-, a,
between
at
b
being
semi-axes
36.
of the
ellipse.
area
[CLARE, etc.,1882.]
the
curve 5
=
Show
that the
at
in eluded its
atan^,
its
tangent
^
=
0 and
tangent
-
ig V*"" """
tan
-a2 tan
^ + a2 tan "/"
the
of
+ a2log(sec ""
c").
[TRINITY, 1892.]
37. Show p =^isin
38.
that
area
the
Sty and
that
are areas
its pedalcurve the
curve r
=
the epicycloid space between taken from cusp to cusp is ^irA2B.
Show
whose
TT a2(f
+
a(^\/3 + cos^#) has three loops 2\/3), a2(f f\/3), spectively. a\ -far fV") reTT
-
[COLLEGES, 1892.]
39. Find the
area
of
a
loop of the curve x*+y* Za2xy.
=
[OXFORD, 1888.]
40.
Find
the
area
of the
pedal of
the
curve
d*)l, =*("**the
originbeing taken
Find the
area curve
at
x
=
*Ja2 62, y
"
=
0.
one
[OXFORD, 1888.]
of the branches of
41.
included
between its
the
3%2
=
and a2(#2+;?/2)
area
asymptotes.
curve
[a,1887.]
42.
Find
the whole
of the
=
43.
Find
the
area
tf+yi a\x*+y*). of a loop of the curve (mV + n,y)* aV b2y2=
-
[a,1887.]
[ST.JOHN'S, 1887. ]
and find their
44.
areas
Trace
:
"
the
shape
of the
followingcurves,
(i.) (^+^2)3 =aay*.
+ 2/)3 a^?/4. (^2 (ii.)
=
[BELL, etc., SCHOLARSHIPS,
45.
1887.]
Prove
that the 3?
'
area
'
of
V2
1 / X2
V2\2
"
7TC2/
182
INTEGRAL
CALCULUS.
46.
Prove
that
the
area
in
the
positive
is
quadrant
of
the
curve
(av+w^w
47. Prove that the of the
^(5+5).
[a;18900
area
curve
f")
is
-3""
+
(V2
-
a2)
[ST.
tan-1
-
.
JOHN'S,
1883.]
48.
Prove
that
the
area
of
the
curve
9,aV
h
62 where
49. is
c
less that
than the
both
a
and of the
5,
is
7r(ab
"
c2).
+
[OXFORD,
1890.]
Prove
area
curve
^4-3o^3
a2(2^2+y2)=0
TRIPOS,
curve
is
fTrtt2.
50.
[MATH.
that the
areas
1893.]
Prove
of
the
two
loops
of
the
are
(32^
+
24^3)
a2,
and
(167r-24\/3)a2,
[MATH.
TRIPOS,
1875.]
CHAPTER
XI.
SUKFACES
AND
VOLUMES KEVOLUTION.
OF
SOLIDS
OF
153. It
was
Volumes shown about
x
=
of Revolution
in
about if the
the
curve
a"axis.
Art.
5
that
x
y=f(x)
the the
revolve ordinates formula
the
axis
x
=
of
x2
the is to
x^ and
portionbetween be obtained by
*2
Tr2
.
dx.
154. More
About
any
axis.
the
if generally, if PN be
revolution
be about
any
line
a
AB,
and
any
perpendiculardrawn
from
184
INTEGRAL
CALCULUS.
point P on the curve upon the line the volume contiguous perpendicular,
or
AB is
and
P'N'
a
as expressed
if 0 be
a
givenpointon
the line AB
155. Ex.
1.
curve
Find
the volume
formed
by
the revolution
of the
loopof
Here
the
f=x2?"^ (Art.130, Ex. =/
J
o
3) about
the tf-axis.
volume
I x*a~xdx. 7ry2dx=7r J
J
o
a
+ x
Puttinga +x=z,
this becomes
rf2a3 log
Ex. 2. of a Find the volume
z
-
5a2z
+
2az2
~
3 _J"
arc parabolic
formed tion of the spindle by the revoluthe vertex to one about the line joining
extremity of
the latus rectum.
Fig.
Let the Then and be parabola
40.
y2
is y P
=
4o
the axis of revolution
=
2^7,
"fi
VOLUME
OF
REVOLUTION.
185
Also
and
volume
=
.
dAN
o
4?r
75
156.
Surfaces
of Revolution.
the curved
Aain, if S be
out
surface of the solid traced
arc
y the revolution
of any
AB
about
the
^c-axis,
Fig. 41.
suppose
PN PN, QM two adjacent ordinates, being the of the smaller,3s the elementaryarc PQ, SS the area
zone elementary
traced
out
by
the revolution
of
PQ
186
INTEGRAL
CALCULUS.
about
the
#-axis, y and
it
y +
as
Sy
the
lengths of
that the
the
ordinates of P and Q. Now take we may traced would from
out
axiomatic is
at the
area
by PQ
in its revolution
greaterthan
distance PN
at
it
be if each
point of
it
were
the axis, and less than distance QM from the axis. Then therefore
if each 8s and
pointwere
a
SS lies between
in the limit r/^ -j=
^y
have 8
and 2w(y + Sy)Ss,
we
f
2-7T2/
as
or
"
\
This may
be written
as
may
happen
to
be of
convenient
-r-"
in any
particular
example,the
from
values
beingobtained -j-" -^, etc.,
the differential calculus.
1.
157. Ex.
formed Here
Find
the
surface of the
of
a
belt
=
of the about
paraboloid
the #-axis.
by
the revolution
curve
y2
"ax
=
dx
V x
dx "/"
y"dx /X"1
dx
Ex. line. Here
2.
The
curve
r
=
a(l+
=
cos
6}
revolves
about
the
initial
Find
the volume
=
and
surface
TT
of the
formed. figure
volume
/try^dx
TT
/?'2sin20 d(rcos 0)
=
/a'2(l 0 -f cos2$), cos -{#)2sin2#a c/(cos
188
2. A
INTEGRAL
CALCULUS.
its chord.
Show
of radius a, revolves round quadrant of a circle, that the surface of the spindle generated
and
that its volume
=
-^-(10 3?r).
-
3. The rectum
curved
part of the parabolayL "ax cut off by the latus revolves about the tangent at the vertex. Find the surface and the volume of the reel thus generated.
=
THEOREMS 158. I. When line in its own any
OF
PAPPUS closed
OR
GULDIN. revolves about
a
curve
plane,which does not cut the curve, the volume of the ring formed is equal to that of whose height and a cylinderwhose base is the curve is the length of the path of the centroid of the area of the curve.
Let the #-axis be the axis of rotation.
area
Divide
the
(A)
with
elements up into infinitesimal rectangular sides parallel such as to the coordinate axes,
Fig. 43.
each PjPgPgP^,
of
area
SA.
Let the ordinate is
Let rotation take placethrough an 89. Then the elementarysolid formed and
PlNl y. infinitesimal angle
=
on
base
SA
its height to first order infinitesimals is ySO,and therefore to infinitesimals of the third order its volume is SA
.
THEOREMS
OF
PAPPUS.
189
If the
rotation
be
through
. .
any the
finite
whole
angle a
area
we
obtain by summation SA y If this be integrated over have for the volume curve we
a.
of the
of the solid formed
a!i/cL4.
Now of
a
for the ordinate the formula of masses number m2, ..., mv
X?7? II
of the centroid with ordinates of
2/i" 2/2'""""
is y=
-^ then
y
"
we
seek
the value
of the curve, each the ordinate of centroid of the area element 8A is to be multiplied by its ordinate and the sum of all such productsformed, and divided by the
sum
of the elements,and
we
have
or
in the
language of
the
Calculus Integral
A (yd (yd A
J
y
=
_
=
i
-
.
\dA
Thus Therefore But is the
A
A
volume
is the
formed of the area
=
A(ciy). revolvingfigureand
ay
of the length
path
of its centroid.
This establishes the theorem. COR. If the curve perform have and form a solid ring, we
a
=
a
complete revolution,
=
2-7T and
volume
A(2jry).
closed curve revolves about a 159. II. When any line in its own plane which does not cut the curve, the curved surface of the ring formed is equalto that
190
INTEGRAL
CALCULUS.
and whose of the cylinderwhose base is the curve is the lengthof the path of the centroid of the height perimeter of the curve. Let the #-axis be the axis of rotation. Divide the such as PXP2 s up into infinitesimal elements perimeter each of lengthSs. Let the ordinate PlNl be called y. Let rotation take place throughan infinitesimal angle S9. Then the elementary formed is ultimately area a with rectangle of the second sides Ss and ySO,and to infinitesimals order its area is Ss y"9.
.
Fig. 44.
If the rotation be through any finite angle a we obtain by summation Ss ya. If this be integrated the whole of over perimeter the curve have for the curved surface of the solid we formed
.
an/cfe.
of the seek the value of the ordinate (rj) centroid of the perimeterof the curve, each element of Ss is to be multiplied by its ordinate,and the sum
we
If
THEOREMS
OF
PAPP
US.
191
all such
formed, and products
we
divided
by
the
sum
of
the elements,and
have
Lt
or
in the
languageof
the
Calculus Integral
\yds ^yds
n
\ds
Thus and the surface formed
s
=
s
\yd8=8tj,
s(afj).
is the
of the revolving and perimeter figure, arj is the length of the path of the centroid of the perimeter. But This establishes the theorem. Con. If the curve perform and form
a a
a
=
completerevolution
2?r and
solid
we ring,
have
=
surface
Ex. The volume of
2 -73-77). s(
and surface of an formed anchor-ring by radius about of line in circle the a a a plane of the circle at distance d from the centre are respectively the revolution volume surface
=
Tra2X 2?rc? 27T2a2o?,
=
=
2:ra
x
Zird
=
4ir2ad.
EXAMPLES.
1. An
revolves ellipse Find revolves of the
about
major
2. A
an
axis.
the volume about
the tangent at the of the surface formed.
a
end
of the
square formed. scalene does not
extremity
3. A
other
to a diagonalthrough parallel diagonal. Find the surface and
volume
which
trianglerevolves about any line in its plane the cut triangle. Find expressionsfor the
of the solid thus formed.
surface and
volume
192
INTEGRAL
CALCULUS.
160. When
Revolution
any
we
of
a
Sectorial
area
Area. revolves about the
sectorial
may
OAB
the
initial line
divide
area revolving
infinitesimal
area
sectorial elements denoted
such
as
up into OPQ, whose
may
be
to first order
infinitesimals
by
|r"2o0.Being ultimatelya
centroid is f of the way in a complete revolution from
element, its triangular 0 along its median, and
travels
a
the centroid
or
distance
r sin 6) 27r(f
f irr sin 9.
Fig. 45.
Thus
by
Guldin's
first theorem is
the volume
traced
by
the revolution
of this element
to first order
and infinitesimals,
therefore
area
the volume OAB is
traced
by
the revolution
of the whole
f
161. If
x we
=
sin 9 d9. 7r[r3si
we
put
y
=
rcos9,
rsin9,
=
and
"
~
=
have
=
r3sin 9 S9 r3sin 9
.
r3sin
lf) (9$(tan
=
7
=
r*cos*9t St
xH
St,
EXAMPLES.
193
as expressed
and the volume
may
therefore be
(xHdt.
EXAMPLES.
and surface of the right the volume 1. Find by integration circular cone formed by the revolution of a right-angled triangle about a side which contains the right angle.
2. Determine
generatedby
3. Prove
the entire volume of the ellipsoid which is the revolution of an ellipse around its axis major.
[I.C. S.,1887.]
that the volume of the solid generated by the round revolution of an its minor is a mean axis, ellipse portional probetween of the those generatedby the revolution and of the auxiliary circle round the major axis. ellipse
[I.C. S., 1881.]
that the surface of the prolate 4. Prove formed spheroid by the revolution of an of about its e ellipse eccentricity major axis is equal to
2
.
area
of
ellipse
.
formed Prove also that of all prolate spheroids of surface.
an
by
of ellipse the volume the
curve
given
area, the
sphere has
the revolution the greatest of
C. S., 1891.] [I.
5. Find
of the solid
producedby
the revolution
x.
the
loopof
y^"x^
about
the axis of
C. S., 1892.] [I.
6. Find revolution
the surface and volume of the reel formed of the cycloid round a tangent at the vertex
by
the
7. Show that the volume of the cissoid y2(2a
"
of the solid formed tion by the revolu^)=x3 about its asymptote is equal
to 2?r2a3. 8. Find
1886.] [TRINITY,
the volume of the solid formed by the revolution of C. S. 1883. ] (a x)y2 a?x about its asymptote. [I.
-
the
curve
=
,
9. If the
curve
r
=
a
+ b
cos
0 revolve
about
the
initial
line,
be
show
that the volume greaterthan b.
E. i. c.
4- b2) provided a generated is "7ra(a2
[a,1884.]
N
194
INTEGRAL
CALCULUS.
10.
Find the
the
volume radius
of of
the the
solid
formed of the
by
curve
the r^
revolution
=
about between
prime
6
=
loop
0
and
0=|.
if the
area
[OxTOKD,
1890.]
11.
Show
that
lying
within
the
cardioide
and
without
the
initial
12.
line, the
The
parabola r(l+cos $) 2a, volume generated is 187ra3.
=
revolves
about
the
[TRINITY,
1892.]
the
loop
line y=a.
of
the Find
curve
Zay2=x(x
volume of
"
straight
13.
area
the
about a)2 revolves the solid generated.
[OXFORD,
Show of that the bounded coordinates of the centroid of the
1890.]
sectorial for its
r=f(0)
by
the
vectors
0=a,
6
=
ft,has
coordinates
f
14.
Show initial
that line
the
at
centroid distance
of
the
cardioide the
r
=
a(l"
cos
$)
is
on
the
a
from
-
origin.
round
is
6 15.
If
the
"
cardioide
r
=
a(l
the
+
"
cos
#)
revolve
the
line
p=rcos(9
y\
prove
that
volume
generated
y.
3^7r%2
16.
f 7T2"3cos
[ST. JOHN'S,
1882.]
The
a
about volume
is very 0\ where small, r=a(l -ecos e the initial line. Prove to parallel tangent of the solid thus generated is approximately
curve
revolves that the
27r2a3(l+e2).
17. the The lemniscate Show that r2 the
=
[I. C. S., 1892.1
about
_
a2cos2#
volume
revolves
a
tangent
at
pole.
generated
is
-
196
area
INTEGRAL
CALCULUS.
RSTU is Sx.Sy, and its mass rectangle be regarded (to the second order of smallness) may as 0(0,y)Sx Sy. Then the mass of the strip PNMV may be written
of the
or
in
conformitywith
the
notation
of the
Integral
Calculus
y =f(x). In performing this integration (withregard to y) x is to be regarded
=
between
the limits y
we
0 and
as
constant, for
masses mass
the
i.e. the
of the limit of the sum finding of all elements in the elementary PM, strip of the strip PM.
are
If then all such lie between be written
we
search
for the
mass
of the
area
AJKB
as strips
be summed which the above must the ordinates AJ, BK, and the result may
which
may
be written
with the limits of the integration b. from x a to x
=
regard to
x
being
=
DOUBLE
INTEGRATION.
197
Thus
mass
of
area
A JKB
=
or
n
164. Notation. This is often written
ff "j"(x,y)dxdy,
order. the elements dx, dy being written in the reverse There is no uniformly convention as to the accepted order to be observed, but as the latter appears to be shall in the the more used notation, we frequently volume adopt it and write present
'x, y)dxdy
when with
we are
to be made to consider the first integration y and the second with to regard
x,
to regard
and
when the firstintegration is with is to say, the right-hand element
regardto
x.
That
indicates the first
integration.
If the surface-density of a circular disc bounded by xP+y2 a2 be given to vary as the square of the distance from the y-axis, find the mass of the disc. "JL^ ^ Here we have [juv2 and its of the element 8x 6y, for the .mass is therefore /*#2"#6y, and the whole mass will be mass Ex.
=
//
The limits for y w411 be ;?/ 0 to y=*Jdl xL for the positive and for x from #=0 The result must then to x=a. quadrant,
=
"
198
be
INTEGRAL
CALCULUS.
multiplied by 4, for the four quadrantsthe the first quadrant.
the
mass
distribution of the whole
being symmetrical in
is four times that of
Fig.47.
Putting x=asm0
and
dx=acosOdO,
we
have
165. The
Other
same
Uses theorem
of Double may be
Integrals.
used
purposes, of which we givea which may the field to indicate to the student serve of investigation But our now scope in open to him. the presentwork does not admit exhaustive treatment of the subjects introduced.
for many other few illustrative examples,
DO
UBLE
INTEGRA
TION.
199
the
Ex.
Find
the statical moment
r2
'
of
4,2
=
a
quadrant of
ellipse
_+"_ 9 1,9 a2 62 about the
!
the surface-density beingsupposeduniform. y-axis, Here each element of area 8x8y is to be multiplied by its surface density in the constant is case cr (which by hypothesis and the the and its distance from sum x supposed) by y-axis, the whole of such elementary quantities is to be found over will from The be limits of the quadrant. integration y=0 to
7
y
=
_Va2
a
"
x* for ?/ ; and
from
#=0
to
x=a
for
x.
Thus
we
have
moment
=
/
00
/
Wa2 crxdxd'u=^"\
a
"
x2dx
J J
)
0
_"rbr _(a?-x2)%-\a_o-ba*
aL
3
Jo
3
166. The
at
Gentroids. formulae
Cartesians.
in statics for the coordinates
proved
a
of the centroid of
number
of
masses
mv
m2, m3,
.
.
.
,
are etc., y2), (xv y^, (x2, points
"
_ ~
apply these to find the coordinates of the centroid of a given area. (See also Arts. 158, 159.) be the surface-density For if at a given point,
We
may
o-
then
or
Sx
Sy is the
mass
of the
element,and
-
_ "
ox 8y)x S(cr I("rSxSy)9
or,
as
it may
be written
when dx
the limit is taken
I dy \\"
arx
\ardxdy J
200
INTEGRAL
CALCULUS,
jja-ydxdy
Similarly
the limits of
summation
~
ff~
j J
\\a-dxdy
integration beingdetermined
will be effected for the
so
that the
area
whole
in
question.
Find
Art.
the
centroid
of the that the
quadrant elliptic
the limit
of the
Example
of the
in
165.
was
It
proved
moments
there about
of the
sum
mentary ele-
was y-axis
?"
"
.
Also Hence
/ /"rdxdy=
*="
mass
of the
quadrant=^^-.
STT
3/4
"
=-"
Similarly
y
=
167. Moments When
square
of Inertia.
the sum Inertia with regardto the line. Such quantities in of greatimportance are
Ex. Find
is multiplied by the every element of mass of its distance from a given line, the limit of of of such products is called the Moment
Dynamics.
the moment of the paraof inertia of the portion bola f/2 4a# bounded by the axis and the latus rectum, about the #-axis supposing the surface-density at each pointto vary the nth power of the abscissa. as Here the element of mass is
=
/x being a
constant,and the
Lt
moment
of inertia is
\ //,
V*"a? 8y 2/i#
are
or
dy, \y*xndx
for
x
where
the limits for y
from
0 to
2vW, and
from
0 to
a.
DOUBLE
INTEGRATION.
201
We
thus
get
In.
=
Mom.
"
3
o
=
" f*
oj
3
U
+
fo
the is givenby parabola
Ca\~
~~l
Again,the
mass
of this
of portion
ny\/ax
Thus
we
l*xndxdy p\ \y\
=
xndx
--
271 + 3
have
Mom.
In. about
0ff=3
EXAMPLES.
the the first quadrant of the circle ,272+^2=a2 Find varies at each pointas xy. density the mass of the quadrant, (i.) " its of gravity, centre (ii.) .("") its moment of inertia about the #-axis. (iii.)
1. In 2. Work out
surface
the
results corresponding axis and
for the
of portion
the
parabolay^=^ax bounded by the varying as xpyq. surface-density
3. Find
the latus rectum, the
varies centroid of a rod of which the line-density the distance from one end. as Find also the moments of inertia of this rod about each end and about the middle point. the centroid of the trianglebounded by the lines at each when the surface-density y mx, x a, and the #-axis, from varies the the of distance the as point origin. square Also the moment of inertia about the #-axis. the
= =
4. Find
168. For of
area
Polar
Curve.
Second-order
use
Element. for
our
polarcurves
a
it is desirable to
element
second-order Let OP, OQ be two
infinitesimal of different form. radii vectores of the contiguous Let
curve
r=/(0);
Ox
the initial line.
0, 9 + SO be
202
INTEGRAL
CALCULUS.
angular coordinates arcs RU, ST, with and let SO respectively,
the first order.
area
the
of P
centre
and
Q.
Draw
two
cular cir-
and
0 and radii r and r + Sr of Sr be small quantities
Then
sector OSTsector ORU
RSTU=
=
r
"9 Sr to the second
may
order,
be considered
and
a
to this order
RSTU
therefore rSO
of rectangle
sides Sr
(RS)
and
(arcRU\
Fig,48.
at each pointR(r,9) is surface-density of the element RSTU is (tosecondthe mass cr "f)(r, 9), order quantities the mass of the sector err S9 Sr, and
=
Thus
if the
is therefore
Ltdr==Q[2o-rSr]S9,
the summation
being for
all elements
from
r
=
0
to
r=f(9),i.e.
"rrdr\80, Q/(0)
-i
in which
and
9 is integration taking the limit of
to be the
sum
regardedas constant,
of the
any sectors
for
infinitesimal values
of S9 between
radii specified
204
INTEGRAL
CALCULUS.
0
or
(Art.164)
=2/
C~% rZacoaO
/
pr*dOdr
169. The
Centroid.
Polars.
a
the sum by finding any of the moments of the elementary masses about that line and dividing of the masses. by the sum Thus its
err
distance of the centroid of line may be found as before
sectorial
area
from
SO Sr
beingthe
r cos
element the
of
mass
and
r cos
0
its moment abscissa,
about
is y-axis
6
.
a-r
SO ST.
rcos(9. a-rdOdr Thus
JJ
x
\o-rdOdr j
r
sin 0
o-rdOdr
dO dr
circle in
and
similarly
'~fl
a-r
the
half of the the centroid of the upper of Art. 168. example established the result for that semi-circle that We Ex.
1.
Find
Also
between
the
limits
r=0
and
r
=
2acos
0 for r, and 0=0
to
/* frcos
^or c^^ dr=
Tfji 0R"
cos
"S
^^
53
15
'
DOUBLE
INTEGRATION.
205
and
I
/rsm6crrd6dr=
I jnsin0
-
dO
Jo
J J
JQ
L4
/~2
sin 0 cos40 dO
Ex.
2.
Find
the
centroid
of
the
area
bounded
by
the
cardioide
being r=a(l+cos #),the surface-density
uniform.
Fig. 50.
centroid is The abscissa we have
on evidently
the
initial line.
To
find
its
/ Ir cos 0 rdOdr
"x"
'rdOdr
r=0 to
the from
limits for
0 to
TT
r
being
from
r=a(l+cos
0),and
for 0
(and
double, to take
=2
in the lower
'"
half).
The
numerator
fT-1
J L3J0
+ 3
cos0o?0
-
cos20 0 1a3 /"'(cos
+ 3 cos30 +
cos*0)dO
206
INTEGRAL
CALCULUS.
=
cos2"9 | T(3
a3
+
cos46
;
'
'
1 ^4-'
'
-
'
-
2 4 o3;r 5
2
5
4
2
2
/TT
L2Jo
0
=
r~r2" Ia(l+cos 6)
dO
a2
T(l +2
0
cos
0+cos20)dO
2a2/ (l+ cos26")d"9
rf r
Hence
x
=
-?ra3
/ -?ra2
=
-a.
varies as the nfh power Ex. 3. In a circle the surface-density 0 of the distance from a point on the circumference. Find the about an axis through 0 perpenof inertia of the area moment dicular to the planeof the circle. and the diameter for initial line, the Here, taking0 for origin radius. is the r=2acos a The density 6, being bounding curve
=p,r".
Hence
moment
Hence
of the element and its rSOSr is //,rn+1S$Sr, axis is //,rn+38$ of inertia about the specified 8r. of inertia of the disc is the moment the
mass
f ffj
where the limits for
r are
0 to 2a
cos
0, and
for
0, 0
to
~
(and
double).
Thus Mom. Inertia
=
n
rcos"+4(9 dO J?^L(2a)"+4 4
+
J
DOUBLE
INTEGRATION.
207
Again, the
mass
of the disc is
r"5"
=
/*2acos0
2|J ^o
frcosw+20d0. _?^L(2a)w+2
n
=
+ 2
Jn
Hence
Mom.
Inertia
=
4
EXAMPLES.
1. Find
the centroid the the
of the sector
of
a
circle
(a) when when (ft)
2. Find
as
is uniform, surface-density varies surface-density
as
the
distance
from
the centre. the centroid of a circle whose the nth power of the distance from a its moments of inertia the tangent at 0, the diameter through 0. varies surface-density
point 0
on
the circumference.
Also
(1) about (2)about
3. Show
of uniform of inertia of the triangle that the moment the a nd lines bounded the ?/-axis by surface-density mlx+cl^ y about is the #-axis, y=mtfc + c"ft
=
Ml
6
GI-%
V
\ml
"
m2J
'
where
M
is the the
mass
of the
triangle.
of triangle uniform
4. Find
moments
bounded surface-density
of inertia of the by the lines
about the
the coordinate
axes
; and
same
show
as
that those
if M of
be the
mass
of
"
triangle, they are the placedat the mid-pointsof
5. Show
equal masses
uniform minor
axes
the sides. of inertia the of
a
that
2
the
2
moments
ellipse
are
bounded
by
---
_--f fC" a2 62 and
1 about and
major and
a 2
respectively
about
4
4
~,
line
I
through the
M
centre
mass
7)2
"
and to perpendicular of the ellipse.
its
plane,M"
,
being the
208
INTEGRAL
CALCULUS.
6.
Find
a
the
area
between
the
circles
r=a,
r=2acos#;
as
and distance
assuming
from the
surface-density
find
varying
inversely
the
pole,
the the
(1) (2)
centroid,
moment
of
inertia
to
about
a
line
through
the
pole
perpendicular
7.
Find for the
area
the
plane.
between the
curves
included
(1)
the
coordinates
of
its
centroid
(assuming
#-axis,
area
a
uniform
surface-density), (2) (3)
the the
moment
of
inertia
about when
the this
volume
formed
revolves
about
the
8.
Find
a
the line for for
moment
of the
inertia
of
the
lemniscate
to
r2
about
through
a
pole
perpendicular
its
plane
(1)
(2)
uniform
a
surface-density, varying pole.
as
surface-density
from the
the
square
of
the
distance
9.
Find
(1)
the
coordinates
of
the
centroid
of
the
area
of
the
cycloid
#=a(0-hsin$),
(2)
the volume formed about about about the the the
y=a(l by
base axis its
"
cos0)
;
revolution
(a)
(y=2a), (#=0),
at
8
tangent
the
vertex.
ELEMENTARY
DIFFERENTIAL
EQUATIONS.
E.
I.
C,
212
DIFFERENTIAL
EQUATIONS.
for the of the
same curve
definite value, the same different for different curves Problems
and
but
family.
in which it is necessary occur frequently to treat the whole family of curves as, for together, each instance,in findinganother family of curves,
member
set at
a
of which that
to be
intersects each member
a
of the former
givenangle, say
for such
not to
manifest letter
a
rightangle. And it will be the particularizing operations,
appear
or as we a
ought
constant
in the be treating of the
functions
one
operated upon,
curve
should
individual
of the
system
"
instead
whole
familycollectively. Now be got rid of thus : a may Solve for a ; we then put the equation into 0(",y)
=
the form
a,
........................
(2)
regard to x, a goes out, and an equation involving x, y and yv replaces equation(1). This is then the differential equation to the family of curves, of which equation equation (1)is the typical
upon of be
case a
and
differentiation with
member. of the differential solve for the
In the formation to impracticable
we
equationit may
In this
constant.
differentiate the
equation
=
f(x,y,a)
with
to respect
x
Q
'
........
.............
(1)
and
obtain
and
thus
then
eliminate
a
between
equations(1) and (3),
x, y, and
a obtaining
relation between
yv
which
is true for the whole
family.
the the lines obtained by family of straight in the constant equation arbitrary
example,consider values to giving special
For
ORDER
OF
AN
EQ UA TION.
213
Solvingfor
and
or
w,
differentiating,
without Blether-wise, first solving for m, we have y= yi
=
m,
and
therefore then is the
y=%yidifferential
This
equationof
expresses
all
straightlines
and passingthrough the origin,
the obvious geometrical line is the same that fact that the direction of the straight as of the same line. of the vector from the origin at all points
172. Again, suppose the of curves the family to be
representative equationof
=
b whose values two containing of the family. A the several members particularize differentiation with regard to x will result in single a relation connecting x, y, yv a, b ; say
ftx9y,a,b) 0, constants arbitrary a,
(1)
If
we a
4(x9y9yl9a9b)0 differentiate againwith regardto
=
(2)
x we
shall
obtain
relation
connecting x,
2/2"a" V" 2/i"
y, yv y2, a, b ; say
\MX
and from
") ""
=
(3)
will
b may cally theoretia and equations appeared be eliminated (if they have not alreadydisand there by the process of differentiation), result a relation connecting x, y, yv y2 ; say these three
the
="" F(x"y,yi"yz) differential equation of the family.
of
an
173. Order
Equation.
We define the order of a differential equation to be the order of the highest differential coefficientoccurring
in it ; and
we
have
seen
that
if
an
equationbetween
214 two
DIFFERENTIAL
EQUATIONS.
unknowns contains one arbitraryconstant the result of eliminating that constant is a differential equation of the firstorder; and if it contain two
constants arbitrary
the result is And
n our
a
differential equation
of the second
so
that
to
order. to eliminate
: argument is general shall constants we arbitrary
have
a
and the result is to n differentiations, proceed an(l differential equationconnecting x, y, yv ...,2/n" order.
c
is therefore of the nth
Ex.
1.
Eliminate
a
and
from
the
=
equation x2+y2=2ax +
c.
x -f yy^ a. Differentiating, Differentiating again, l+^+y^^^ and the constants we having disappeared
have
obtained
as
their
differential equation of the second order (?/2 being differential w hich the highest coefficient involved), belongsto all
a
eliminant
circles whose Ex.
2.
centres
lie on
the #-axis.
the differential equationof all central conies Form whose coincide with the axes of coordinates. axes the typicalequation of a member of this family of Here
and and
we
have
whence is the differential
x(y? + yy2) -yyl=0 sought. equation
an
174. Elimination Now
irreversible process.
this process of elimination is not in general a reversible process, and when wish to discover the we of a family of curves equationof a member typical when the differential equation is given, are pelled comwe of integration, to fall back, as in the case upon
cases,
a
set of standard
are
and
many
equations may
arise
which We
not solvable at all.
to solve however, that in attempting may infer, differential equation of the nth order we to a are search for an algebraical relation between x, y, and n
VARIABLES
SEPARABLE.
215
these constants arbitraryconstants, such that when eliminated the given differential equation will are result. Such is regarded as the solution most a general solution obtainable. DIFFERENTIAL 175. CASE All all the There
are
EQUATIONS
OF
THE
FIRST
ORDER.
five standard
forms.
I. Variables
Separable.
which it is
equationsin
x's to
come one
other,
under
to get dx and possible side,and dy and all the y's to the this head, and solve immediately
by integration.
Ex.1, have Thus if
sec
y=
sec
x-",
dx y
we
cos
x
dx
x
"
cos
dy,
,
and
a
integrating, relation containing one
Ex.
2.
sin
=
sin y + A
constant arbitrary
x
=
A.
If
xy-^*
dx
+ y)dy, (y2
32
y+l
we
have
(x
\
+
2
-
dx )
J
=
x
and
therefore
"
+
logx"^
32
+y~+ A,
A.
2
one containing
constant arbitrary
EXAMPLES.
Solve the
y
I
differential equations : following
"
1.
dy=x*+x+\
'
dy
'
y*+y+l
in Ex.
3
dx
*++l'
that member
every
dxx*+x+l
member of the of the set in Ex.
/
4. Show
cuts
every
2 at
family of curves right angles.
216
7. Show
DIFFERENTIAL
EQUATIONS.
the square
vector
are
that all the
curves
equal
to
square
for which of the radius
of the normal is either circles or
hyperbolae. rectangular
8. Show
that
a
makes
no
a
constant
for which the tangent at each point angle (a)with the radius vector can belong to
curve
other class than 9. Find the
r=Ae^
cot
a.
of the curves for which equations (1) the Cartesian subtangentis constant,
(2) the
Cartesian subnormal is constant, the Polar is (3) subtangent constant, the is constant. Polar subnormal (4)
10. Find
tangent
the Cartesian equationof the is of constant length.
curve
for which
the
176. CASE
II. Linear
Equations.
the form
of [DEF. An equation
when
P, Q,
.
.
.
,
K, R
are
functions
of
x
or
constants
is
lies in the fact that said to be linear. Its peculiarity differential coefficient occurs raised to a power no higherthan the first.] of the first As we are now equations discussing
order,we
are
limited for the
to present
the
case
If this be
seen
' *
throughoutby multiplied
may write it d
er
it will be
that
we
dPe
Thus
a
,
/Pefccv n
)="^fPdx
"
yefPdx=\Qe/Pdxdx+
between
x
relation
and
differential equation,and It is therefore constant. The called factor
"
'
the given satisfying containingan arbitrary the solution required.
y
e
which
of the
an
a equation factor." integrating
the left-hand member differential coefficient is perfect rendered
LINEAR
EQUATIONS.
=
217
Ex. Here
mav
1
.
Integrateyi+xy e-**
fxd~
or
-
x.
e2 is
an
and factor, integrating
the
equation
be written
d
ax " *2
-
-
(ye*)=xe*,
or
ye*=e*+A,
+
i.e. Ex.
2.
y
=
l+Ae
2.
Integrate
^l +
dx
-y=x2.
x
Here may
the
factor integrating
is e
Jldx =elogx=x,and
x
the
equation
be written
*JH"-+
and
xy=--+A,
x*
177.
Equations reducible
if equations,
not
to
linear
form. the linear
Many
form
immediatelyof
be immediatelyreduced may variables. One of the
most
to it
_
by change of
is that
the
important cases
of the
equation
Or
y-n
Putting
we
yl~n
=
z,
have
y-^dy=^
218
DIFFERENTIAL
EQUATIONS.
or
which
and is linear,
ze
its solution is
(l-ri)fpdx
=(1" ft) \Qe
=(1
,.,
"
,~
\fr\ Q-~n)fPdxJ
dx+A,
dx+A. A
l-n
(\-ri)fPdx
e
i.e.
y
We ri)
v
f^ (l-ri))
x-,
Ex.1,
Integrate-^ + ^=?/2.
1; ^-2^+^
=
Here
or
putting
-=0,
t7
dx and the
x
factor being integrating -fix*
ej*
=e-loex
we
have
^(^=-1,
dx\x)
x
i.e.
?=logi
X X
i.e.
-=Ax y
"
Ex. 2.
the equation-jf. + x sin 2y Integrate
Cv"^7
=
we Dividingby cos2y
have
tan
ec2y-^ + 2#
dx
y"x^.
Putting
we
tany=2,
have
^
GW?
+
2^=^,
and
the
factor integrating
is
"J2xdx Or e*z, giving
220
DIFFERENTIAL
EQUATIONS.
16.
Find
sum
the of the
polar
radius of the the square class whose
equation
vector
of and
vector.
the the
family polar
of
curves
for
which
as
the the
subnormal
varies
nth 17.
power Show
as
radius
that the
curves
for the
which
the
radius upon
of the
curvature
varies
of
perpendicular equation
A is
normal
belong
Jc
to
a
the
pedal
and
r2-p2=^
*
+ %*
"^
being
18.
given
constant
arbitrary.
Integrate
the
equations
CHAPTER
XIV.
EQUATIONS
OF
THE
FIRST
ORDER"
CONTINUED.
HOMOGENEOUS
EQUATIONS.
CLAIRAUT'S
ONE FORM.
LETTER
ABSENT.
178.
CASE
III
Homogeneous
in
x
Equations.
and y may
Equations
homogeneous
be
written
dx if for
(a)
obtain
In
this result
case
we
solve form
possible
-^, and
a
of
the
Putting
we
y
=
vxy
obtain
v
+
x^
=
"j"(v)"
dv
_dx
'
~"p(v)"v
and
comes
x
the
variables under Case
are
separated
as
and result
the
solution
thus
I., giving
log
Ax
"
r 1
222
DIFFERENTIAL
EQUATIONS.
or
(6) But
solve
we
if it be
inconvenient
"
to impracticable
p
for
-". we
dx
solve for
and
.
write
for
"-
and dx
.
x
have y
=
x"f"(p)
...............................
(1)
with respect to Differentiating
x,
or
dx="["'(p)dp
x
-"
this equation have x expressed as we a Integrating function of p and an arbitrary constant (2) Ax=F(p)(**y) Eliminatingp between equations(1) and (2) we obtain the solution required.
.........................
Ex.
1.
Solve
(x*+y*)ty-=xy.
dx
and
putting
y=v%,
^+v dx
dv
or x"
=
-
dx
or
og=-2
a;2
or
Ay^eW.
Ex.
2.
Suppose the equation to
x
be
dx
\dx)
'
HOMOGENEO
US
EQ
UA
TIONS.
223
Then
p
=
(p +p2)
+
x(l
+
2p),
p
giving
log J,#+2logp
"
-=0,
P
i.e.
and
the
jo-eliminant
between
p2+p="
x
\
and is the This solution eliminant
Axp*=" sought.
is
"
But
when of
it p,
is
or
elimination
algebraically impossible to if performed, the when,
the
as
perform
result
two
the
be
will
to leave manifestly unwieldy, it is customary to regard them containing p unaltered, and would equations whose jo-eliminant if found
equations
simultaneous the
be
required
solution.
EXAMPLES.
Solve the differential
equations
.
.=.
dx
x+y
2.
224
DIFFERENTIAL
EQUATIONS.
179. A
The
Special Case.
~ "
-
dx ax+oy+c the homogeneousform thus : Put x
"
equation
"f
^-
,
is readilyreduced
to
TVi
^
"
a^+ by+ (ah+
so
bk +
'
c)
dg-a'g+b''
Now choose h, k that
^e.
so
that
.1
"
r-/ oc
^"be
=
-"
"
,-
-
-^
ao
"
^
a
ca
"ca
6
Then This
^=
now
equation being homogeneous we may variables an(i ^ne are as separable put n~v^
shown. is 180. There cannot be chosen
before
one as
however, in which when above, viz.,
case,
a
_ ~~
h, k
b 6'
c
a' Now Then let
"
c''
=m a
and
dy
Tx=
n
-~
so
that
--
"
dx
_
7
a "*
=
-
V
-
-
my
+ +
/)
c
drj (am
dec""
and
-,
-
+ ad b)rj
mrj +
c
+ 6c
+bc (am-\-o)r)-\-ac
,
,
".
.
-,
n.
HOMOGENEOUS
EQUATIONS.
225
the integration beingnow separated, be at once performed. may 181. One other case is worthy of notice, viz., ax + by+ c dy dx~ "bx + b'y + c" when the coefficient of y in the numerator is equalto that of x in the denominator with the opposite sign. For then we may write the equation thus + b(ydx+x dy) (b'y + c')dy (ax+ c)dx differential equation exact an being ; the integral ax2 + 2cx + 2bxy b'y2 + 2c'y + A A. beingthe arbitrary constant. variables
_
The
=
"
"
=
,
Ex.l. Put
Integrate =ydx
x+y-Z k, so
that
#="+
k, y
=
ri+
Choose
h and
Icso
that
=
then Now then put 77=0(1,
_
~
l+v
v+1
'
"-
1)2-
l
where
E. I. c.
"=#"1
and
p
v=^~
.
x"\
226
DIFFERENTIAL
EQUATIONS.
*+*
Ex. 2.
Let
f Integrate
dx
=
.
x+y
"
\
.#+y=??, then
.=
..
dx
??
1
if] 1
"
'
and
^
where
?7=
EXAMPLES.
the equations : Integrate
dy
_
dx
bx+ay-b
"
1*
^
8 9. Show
that
a
particle #,
y which
moves
so
that
~
will
always lie upon
that
a
conic section. of the
10. Show
solutions
must
generalhomogeneous
families of
tion equa-
fUL
'
\"
dx)
~)
always represent
similar
curves.
11. Show
that solutions
a a
of
/(-, -j-} J
are
homogeneous in
curves
x,
\X
CLX
of y and some power the typical equationof in x, y and some
that constant, and conversely single member power of a of one
if
family of
constant,the
be homogeneous differential
228
DIFFERENTIAL
EQUATIONS.
x, i.e.the absent
Differentiate with The P
regard to
letter.
and
dx
Thus After the
x
is performedwe integration and this equationand 2/ "/"(p) between of the givenequation is obtained.
=
eliminate p the solution
183. y B. which
absent. the differential
Suppose y
absent from then takes the form
equation,
Since
-^
ax
fj 1J
=
"
this may
be written
ax
dy
dx'
and Thus
therefore if y
variable the
regardedas the independent remarks foregoing apply to this case also.
be dx
we
convenient result of the form
if (i.)
solve
for
-^-, ^
and
obtain
a
dx
,
a5T*S
dx
then
7 dy
=
.
"7-^,
and the
is integral
dx
ONE
LETTER
ABSENT.
229
But (ii.)
if this solution
for
and
dy
solve for
x
-7-
be inconvenient
a
or
we impossible
obtain
yy
/v"
result of the
Then
form
x
=
where (j)(q)
q stands
for
-j-
tiating differen-
with
regardto
y, the absent letter ,
Thus
and After
we integration and equation and x "f)(q). equationis obtained.
=
the
eliminate the
between solution of the
q
this
given
absent
The
or
student
should
note
~
that in either case,
x
y
we absent,
solve for
if possible. by preference
ax
inconvenient solve we or impossible for the remainingletter and differentiate with regard the absent letter to the absent one\ thus considering in either case the independentvariable. as
Ex.
'
But
when
this is
1.
Integratethe equation
Here
dy
and is the solution. Ex.2. Solve dx Then #=
"
2
=
\dx)
x
where
q=.
dy
230
Then
DIFFERENTIAL
EQUATIONS.
absent lettery,
with regardto the differentiating /, 1 \dq l~
*
and and the
# between ^-eliminant is the solution this
equation
and
the
original
equationx=q+~
required.
EXAMPLES.
Solve
1.
the
=
equations:
dy
y +
I.
5.
6.
-B
\dx)
dx
4.
(2a^ + ^2)=a2+2a^7.
dx
8.
=A+".
\dx)
dx
184. CASE
V.
Clairaut's
have
Form,
^=
Writing^9 for
-~-
we
with Differentiating
y=px+f(p) regardto x,
dp
........................
(1)
or
{x+f(p)}0,
either
....................
(2)
whence Now
-^-=0
CLOu
or
cc+//(p) 0.
=
-f
=
"
givesp
=
C
a,
constant.
CLAIRA
UTS
FORM.
231
is a solution of the given differential Cx-\-f(C) C. constant an equationcontaining arbitrary Again,if p be found as a function of x from the equation (3) "+/(,) 0, Thus
y
= =
........................
stillbe satisfied, and if this value of which is the same or p be substituted in equation(1), if p be eliminated between thing, equations (1)and (3) shall obtain a relation between we y and x which also satisfiesthe differential equation Now to eliminate p between
equation(2)will
y=px+f(p)}
0=
x+f(p)I
0 between
is the
same
as
to eliminate
0
=
x+f(0)J
i.e.the
same
=
the line y There are
the process of finding of the envelope Cx+f(C) for different values of 0. therefore two classes of solutions, viz. :
as
(1) The
linear
complete tive," primiconstant. an containing arbitrary
" "
called the solution,
"
(2) The
solution or envelope singular containing and constant from derivable not no arbitrary the complete primitive by putting any numerical value for the constant particular
in that solution.
The
geometrical relation between
these
two
tions solu-
is that of a familyof lines and their envelope. It is beyond the scope of this book to discuss fully the theory of singular and the student is solutions, referred to largertreatises for further information upon the
subject.
232
DIFFERENTIAL
EQUATIONS.
Ex.
Solve
y=jt
JP
By
Clairaut's rule the
is completeprimitive
and
m
the
solution or singular envelope the above
is the result of
eliminating
between
equationand
o=*--2. m2
i.e. The student the will at
y*="ax.
once
a
recognizein
and parabola,
y2- 4a#
=
equation to
the well
solution the singular in the completeprimitive
a
y=mx+"
known
equation of
tangent
to
the
parabola.
EXAMPLES.
and the complete primitive, down Write cases : solution in each of the following
"
find the
envelope
4. y" 5. y
=
(x
"
a)p
"
p^.
6.
(y"
185. The
equation y=x"P(p)+*Kp),
.....................
(i)
may and
with regard to x, by differentiating variable. then considering p as the independent have For differentiating, we
be solved
whence
"
-=
--
dp
which is the solution being linear,
["P(P)*P r .,,Ji eJ"t*P)-P=_ (P) xe"t*P)-P=_
+ tW-Pdp
A
.......
(2)
EXAMPLES.
233
If
now
p be eliminated
between
(2),the completeprimitive of the
will result.
Ex. We Solve have y
p
=
(1)and equations equation original
(1)
2px+p2.
,
.
(2) p^x" %p3 A giving be found from these two equationsmay The jo-eliminant now in equation (2). (1)for p, and substituting by solving equation But if it be an object to present the result in rational form, we may proceedthus : + SA 0,\ (2) 2p3+ 3p2# By equation from (1) 0. / + Ip^x-py j93
" "
..............................
"
=
=
Hen
ce
p*x Zpy
-
-
3A
=
0.
And
between by cross-multiplication
this
equationand
givingas
the eliminant
+ 3Ax)(x* +y) 4(y2
=
(xy
-
3 A)2.
algebraic p being process of eliminating the equations difficultor impossible, in many (1) cases and (2)are often regardedas simultaneous equations but the is the solution in question whose ""-eliminant actual elimination not performed.
186. The
EXAMPLES.
Solve the
equations:
=
1. y=
,
2. y
3. y= 4. y=
p
234
DIFFERENTIAL
EQUATIONS.
8.
The OT2 the Find
tangent
is axis the
at
any
point
to
P the
curve.
of
a
curve
meets
the the
axis
Oy
in of
T,
PTto
and
proportional
Ox. Find
tangent
of
inclination
the
[OXFOKD,
of all
curves
1888.]
possess
9.
differential that the
axes sum
equation
of is of the
constant.
which
the
on
property
the
intercepts
Obtain and
made
as
by
the
tangent
coordinate the
curves
the
complete
tion solu-
primitive
the
10.
equation
in
the
tangent,
as
the
singular
question.
curves
Obtain
the the the the
axes axes
for
a
which is
the
area
of
the
triangle
bounded
11.
by
Form of
and
tangent
constant.
differential
equation
the
of
curves
for
which between
the,
the
length
coordinate
portion
is the
of
tangent
Obtain and
intercepted interpret
constant.
the
complete
primitive
12.
and
singular
the
solution. differential
;
A
curve
satisfies when
equation
its
y=p\x"p))
and
also
that^"=0
x=\
determine
equation.
[OXFOKD, 1889.]
of the
IS.
Find
the
complete
primitive
and
singular
solution
equation
dx Show
\
x2"s and
V^yj
y2
=
'
[OXFORD,
", the
1890.]
14.
that
by
putting
equation
is
reduced
Hence
to
one
of down
Clairaut's its the
form.
write
complete
result.
primitive
and
find
its
singular
solution.
Interpret
236 Then
DIFFERENTIAL
EQUATIONS.
yi
=
2/2
=
zj(x)+zf(x); zj(x)+^'(x)+zf"(x).
we
Thus
on
substitution
get
But
+ Pf(x)+ Qf(x) /'(*)
=
0
by hypothesis.Hence
z
an
equationwhich is linear factor is The integrating
or
for zv
is and the first integral
the second integral whence may effected. solution and the
Ex. Solve
L
be at
once
obtained
da? Here Put then makes
dx
y=x
-r^
Hence
and
the
factor integrating
is
e^
/K" 4- 3^
x
**
or
x*e
4
.
SECOND
ORDER
EQUATIONS.
237
j
x*
Thus
~(zlx^)=x^
a*
5
and
z1x2e*=~+A
_*
whence and the solution
z=-\e
*
+ A
J a?
\ "-e
r
,
is required
-=
5
189. CASE A. If
x
II
One
letter absent.
be
absent, let y1=p,
-
and
the
equation
"f"(y, yv y2)
= =
0
takes the form
and
"f"(y, Q, p, p-S-\
\
dy/
is of the first order.
B. If y be the letter
absent,let
yl =p,
*-"
and
"f"(x, yv y%) becomes
and
again is of
Solve
x
the firstorder.
Ex.1. Here
the
equationyy2+#i2=2#2.
So
is absent.
puttingy"=p
and
y2=p^?, have
we
The
factor integrating
is e^
vAy or
y2,
or
p2y2 "y* +
constant
=?/4+ a4,say.
238
DIFFERENTIAL
EQUATIONS.
Hence
or
i. e. Ex. 2. Here Solve
+ A). y* a2sinh(2#
=
l+#i2=#y2#i"
So
y is absent.
putting y-^"p^
dx
or
"
=
-"--"5,
pdp
x
1+jtr
=
i. e.
^.e.
logx
logVl+p2
+ constant.
or
ady" ^Jx*
"
a?
dx,
giving
a
oy=i?^2_"
constants. being arbitrary
and
b
EXAMPLES.
Solve the
1. ^2 2. 3. 4.
=
: following equations
"
1.
6.
1+3^=^2. i+y!2-^29y22=4iyi=
7.
1
_L2
2-
8a
y2+"/i-y=-e
^"/
y#2=#i3-#i-
[OXFORD, 1889.]
5. "3/2
(l+.yM
the
10. Solve
equation (1
y
=
=2/" havinggiven ~#2)^~#(^)
[OXFORD, 1890. ]
that
^
ow?
=
0 when
0.
11. Given
that #2 is
a
value
of y which
satisfies the
equation
find the
completesolution.
[L 0. S., 1894.]
REMOVAL
OF
A
TERM.
239
190. Term. Let
General
Linear
Equation.
more
Removal
of
a
us
next
consider the
equation general
of
x.
where
Q Pv P2, Puttingy vz, we
. . .
are
,
given functions
=
have
y2
whence
=
vz2 + n(n"
2V"
1)
+ v2z, etc.,
,
--
-
2
The coefficientof 0n_i is If then v be chosen so that
dv
v
P
"
"
or
v
=
e
n
the term
zn-i involving if be so Similarly, v equation
will have chosen
as
been
to
removed. ential differ-
the satisfy
the term containing 0W_2 The coefficient of z is
will have
been
removed.
and
if
a
value this
of
v
can
be found
or
will make
vanish,we expression
=
guessedwhich can, by writing
=
zl the
=
rj,
therefore z2 and zn rjn-i, reduce rjly etc., degree of the equationby unity. The student
and
should
notice
that
this
is expression
the
same
in
240
DIFFERENTIAL
EQUATIONS.
member
y
"
form Hence the
as
the if any
left hand solution
v
can
the given equation when omitted, we by writing y can, vz, reduce the degree of the equation.
=
given equation. be found or guessed of is right hand member
and then Z^
"
of the
Y\,
191.
Canonical
case
Form.
In the
of the
equation
of the
second
degree
the will
substitution has the
y
=
e~l ^dxz
above stated reduce the
by what equation to
the
been
given
sometimes
simpler form
of this
But
at
general solution present effected.
"EXACT"
equation
has
not
been
DIFFERENTIAL
EQUATION.
is
an
192. and
can
When be
p
is
"
q.
xp-r~
^
exact
differential.
integratedwhatever by
yq,
y may
be.
For
denoting
=
\xPyqdx
etc.,
Thus
EXACT
DIFFERENTIAL
EQUATION.
q =p
or
241
It will be noticed
cannot
that when be effected. the above
a
"
p
the integrati
By aid of quicklywhether
193.
given
may often see equation is " exact/' For
we
lemma
if all terms of the form xpyq in which p is first removed, we tell at once can frequently whether the remainder coefficient or not.
Ex.
"
q be spectio by in-
is a
differential perfect
#2
Here, by the lemma, #2y5 and
of xy. Hence
a
x?y" are
differential perfect
and obviously is coefficients, ocy^-\-y first
the differential coefficient of integral this differential equation is
obviously
=-
cos
x+A.
194. A A
more
more
General
Test.
for
an
generaltest
"exact"
differential
equation may
whatever
.
be established in the
case general
the coefficients P0,Pv have, provided they be functions of x. For upon
forms
...
,
Pn, V may
have
denotingdifferentiations by dashes, we integration by parts
n-
=PnBysdx
32/2
-
Pn
-
8^1 +
P"n
-
zV
-
I P"'n
-
etc.
Hence
upon
addition it is obvious
that if
p. i. c.
242
DIFFERENTIAL
EQUATIONS.
the
given equation
is
is
exact; and
that
tegral its first in-
Ex.
Is the
+ 1 2x?y2 + SG^2^ + 24#?/ equationcfiyz
=
sin
x
exact
?
Applying
and Thus the
-
the test,we
have
PQ'" 24^ P3 P2'+ PI equationis exact ; and its
-
=
-
72^ + 72^ first
-
24^
=
0,
is integral
or
This
again will
is satisfied.
be
a
differential perfect
if
which
Hence
-
a
second
=
-
will integral sin
x
be
(8#3 43%
or
-f ^4yx
+ Ax
+ +
B, J?,
that the
4^73y+^4y1= sin^+^^
"
which
again be tested. may is third and final integral ^
=
But
it is
now
obvious
cos.*?+
IB
EXAMPLES.
1. Show
that
the
equation
exact, and
2. Solve + ^7/3 3. Write
solve it the
completely.
.
equation
%i
first
6^/2 +
down
+ sin
x(y* %i) +
~
cos
X33/2 !/}
-
=
sin
^
of integrals
the
equations :" following
'(a) (b) (c)
4. Show
an
that
if the
integratingfactor
equation P2y + P^y^ + P0"y2 //, will satisfy the //,,then
=
F admits
of
differential
equation
244
DIFFERENTIAL
EQUATIONS.
Hence
n equation(1) containing arbitrary and therefore is the solution most general to constants, be expected. No more solution has been found. general The portion f(x) is termed the Particular Integral the n arbitrary and the remaining (P.I.), partcontaining constants,which is the solution when the right-hand member of the equation is replaced by zero, is called the Complementary Function (C.F.).If these two partscan be found the whole solution can be at once a
is
solution of
written
down
as
their
sum.
196. Two There
are
remarkable
two
cases
Cases.
these solutions
can
in which obtained. readily generally
be all
(1) When
constants.
the the
Pv P2, quantities
takes equation
Jn-l
...,
Pn
are
(2) When
ri
the form
r7n-2n/
'
2+-'-+^=
F"
av a2, ..., an beingconstants and V any function of x. is readily The solution of the second case reducible, as will be shown, to the solution of an equation coming under the firsthead.
EQUATION
WITH
CONSTANT
COEFFICIENTS
FUNCTION.
"
MENTARY COMPLE-
197. Let us therefore firstdetermine such an equation as
+ a22/n-2+... + 2/n+^i2/u-i the coefficients being constants attention
the solution of
^n2/
=
0,
.........
(1)
of the
; i.e.for the
we present
confine
"
our
to
the
"
determination
ComplementaryFunction
in the first case.
COMPLEMENTAR
Y
FUNCTION.
245
As
a
trial solution
mn
+
put y Aemx, and we have a1mn-1+ a2raw-2+...+an 0
=
=
.......
(2)
Let the roots of this
be equation
...,
774, m2, ms,
mn,
then all different, (forthe present) supposed
are
and all solutions,
y
=
therefore also
......
+...+A (3) nem*x, A^x + A2e^x+ A3em*x is a solution containing constants n arbitrary Av A2, to be expected. An, and is the most general A3,
...,
198. Two If two
=
Roots
roots
Equal. equation (2)
become
equal,say the of first the solution (3)become two terms mx m2, be regardedas a and since A^ + A^ may (A!+ J.2)e"11*,
constant, there single unity in the number (3) is no longer the expected.
Let
Put
us
of
is of
apparent diminution by arbitrary constants,so that most general solution to be
an
examine
this
more
=
closely.
Then
=
A
m1+A. ^x + A 2e(TOi+*X"
97i2 r
h?x2 +
...
~~\
Alem^x-\-A2(^x\ l+hx+-^(Al+ A2)^x A2
we
=
+
AJi
.
I. xem^x+Azhem^~^
+
. ..
rhy?
~\
A^ and and quantities,
terms
Now
are
two
independentarbitrary
express them in
may
therefore
quantities independentarbitrary by two relations chosen at our pleasure. First we will choose A2 so largethat ultimately small may be written "2, A2h when h is indefinitely finite constant. an arbitrary
of two
other
246
DIFFERENTIAL
EQ
UA
TIONS.
will choose A1 so large and of opposite we Secondly, signto A2 that A^+A2 may be regardedas an arbitrary finite constant Bv Then the terms
vanish with h since Aji has been considered ultimately is confinite and the expression in square brackets vergent and Thus be
contains
h
as
a
factor.
=
A^^+A^e11^ may, when m2 mv and therefore ultimately by B1emiX+B2xemiX" replaced in the whole of arbitrary constants the number
therefore
case.
the terms
solution remains we n, and the generalsolution in this 199. Three
have
obtained
Equal Roots.
the become
case
Consider
next
equation(2) have already been terms, Alem"+A#m*x+A2f?ri*x, by (Bi+B^e^+Atf"**.
=
=
three of the roots of The equal,viz.,m1 m2 m3. when
placed re-
Let
Then
ms
=
mx + h
/
fcZftZ
1 + kx +
we
\
A^x
=
AjPtifc A^x(
=
-^-
+...)"
Thus
for A+Aw
+ Ae'W
have
and
we
may
so
choose
AB,52,and Bv
that
Ov C2,03 being any
constants, whatever arbitrary
k
COMPLEMENTARY
FUNCTION.
247 But
may
be, providedit be
finite
not
absolute
zero.
AJc2
series within being chosen a the square brackets beingconvergent, it is clear that when k is indefinitely diminished, the ultimately,
and quantity, the
form limiting
of this
is expression
200. In
a
Several similar
Roots
manner
Equal.
it will be
obvious
that
if p
roots of the
equation(2)become
m^ m2 loss of
= =
. . .
viz., equal,
=
mp,
there
we
will be no substitute the
for the
in our solution if generality expression + KjxP *)"*", (%!+ K^x + Kfl?+ portionof the complementary corresponding
-
.
.
.
function, viz.,
A^x
More
+
+...+ Apew*"x. A2em*x
201. Generalization. if generally,
be the
ential complementaryfunction of any linear differequationwith or without constant coefficients, what is to replace this expression to re-tain the so as when generality mx m2 ?
=
Let Then
m2
and the terms
become + A2"p(m2) Al"f"(m1) h2
Now
two
putting A^A^ Bly AJi "2) finite constants,the remainingterms arbitrary
= =
248
DIFFERENTIAL
EQUATIONS.
in
when we ultimately disappear approachthe limit diminished. which h is indefinitely Thus Al"f)(ml)+ be replaced J.20(m2) by may
number the same (n) of arbitrary retaining constants An B1952,A2, A^ in the complementary function as it originally possessed. And as in Art. 200 we proceedto show that if may viz. ^i1 m2= p roots become equal, =mp, the terms + 420(m2)+ +Ap"j"(mp) J.10(ra1) be replaced by may thus
...,
=
... " . .
when The
the
of generality
the solution will be retained.
results of Arts. 198, 199, 200 are of course ticular parof this, emix. the form of "p(/m^) cases being
202. When
Imaginary Roots.
a
root of
of (2) equation
Art. 197 is imaginary, real
it is to be remembered that for equationswith coefficients imaginaryroots occur in pairs. for instance, we Suppose, have
where Then
i
=
\/
"
1.
the terms
A^+A^e"**
may
or a
A^
thus
:
"
be thrown
into
real form
=
( Al +
bx bx)+ A2eax(cos bx bx + ( A1 ^2"6a*sin A2)eaxeos
1
sin
"
i
sin
bx)
-
sm
bx,
COMPLEMENTARY
FUNCTION.
249
where
the two
A^+ A2 and Let B^ p
=
=
constants Bl and B2 replace arbitrary (Al A2)irespectively. then cos a, B% p sin a,
"
=
JB*
+
"22
and
=
a
=
tan
-
ijgF.
Then
bx "2sin p cos(bx a). We may thus further replace bx by CLeaa!cos(6aj jB^cos bx + B2eaxsin constants. where C^ and 02 are arbitrary
^cos
fr#+
"
203. For
Repeated Imaginary
Roots.
repeatedimaginary roots we may proceedas that when before,for it has been shown 7772 ??i1, by (J^+J?^***, and Alem^x+A^x may be replaced if m4 by m3, A^X+A^X may be replaced
= =
If then mx
=
m2
=
a
+ ib and
m3
"
m4
=
a
"
f
6,we
may
replace by
that is
(
by
-
sin te] bx + (Bl- B3)i + 53)cos eax[(Bl 60?+ (B2 "4)^ sin 6 + xeax[(B2 + ^4)cos and therefore by 6^+ (72sin 6x+ (74sin +cceaaj((73cos e^CC^cos 6aj) that is by 6aj+ ^ + cc(73)cos tf*(Ci which is the same or thingby
Any
of the last three
which
forms
contain thus
four
constants constants
replacethe Av A2, A^ A^ and
four original
arbitrary arbitrary
the
retain intact
250
DIFFERENTIAL
EQUATIONS.
constants (n) of arbitrary requisite proper number to make the whole solution the most generalto be be extended expected. And this rule may obviously to the case when of the imaginaryroots any number
are
equal.
204. Ex.
Here
our
1.
Solve
the
equation^dx2 is y=Aemxy and
dx
we
trial solution
obtain
whose
roots
are
1 and
"
2.
Accordingly y
and
is the
A^e*and
y
=
are A2e2x
both
solutions, particular
y=A1e*+A2e2x
solution containing constants. two arbitrary general Solve
Ex.2.
-*V-a?y=0.
aOC
"
Here the auxiliary equation is w2 and the generalsolution is
a2
=
0 with
roots
m"
"a,
or
as
it may
be written y
=
(if desired)
ax .Z^cosh
+
^sinh
ax
by replacing Al by
Ex.3. Here Hence Solve
Bi+B*
2
and
A2 by
B^~B
2
the auxiliary equationis m2-f-a2=0 the general solution is y
=
with
roots
m"
+ai.
AjCosax
+
^t2si
or, which
is its
equivalent,
y
=
Bl
+
Ex.4
Solve
?-4|
ax?
ax
5-2y
ax
=
0
or
(D- l)2("-2)y 0,
=
where
D
stands
for
"
.
ax
252
DIFFERENTIAL
EQUATIONS.
3-
4-
5.
6.
S-9 S~3 g=y. g=y.
11. 12.
9.
10.
("-
THE 205. of such
PARTICULAR
INTEGRAL.
function Having considered the complementary V where F(D) stands for an as F(D)y equation
=
av
we a
a2,
...,
an
beingconstants,and Fany
our
function of
of x,
next most
turn
attention
to the mode
and particular integral,
and We
propose useful of the processes write the above
1
obtaining to givethe ordinary adopted.
2/
=
may
equation as
an
where (or [/(D)]F),
^7^
is such
operatorthat
206. "Z""
satisfies
the
fundamental
laws
of
Algebra.
It is shown
in
the
Differential
Calculus
that
the
operatorD
satisfies (denoting -y- j Distributive Law of
viz. Algebra,
(1) The (2) The
Commutative
Law
=
as
far
as
stants, regards con-
i.e.
D(cu)
c(Du}.
PARTICULAR
INTEGRAL.
253
(3) The
m
Index
Law, i.e.
integers. beingpositive Thus the symbol D satisfiesall the elementaryrules the with of algebraical of combination quantities with regard to that it is not commutative exception
n
and
variables. It therefore has a identity analogue. Thus
follows
algebraical symbolicaloperative corresponding binomial theorem since by the
that any
'
rational
7? l
(T)
"""
+
\ JL
.
~Li
2i
we
have
by
an
analogoustheorem
further
may
be inferred without
which operators proof
for
JL
.
"
"
207.
Operationf(D)eax.
a
It has been if r be
proved in the positive integer,
define the
Differential Calculus that
Let
us
D~r operation DrD~ru
=
to be such that
u.
and we shall an integration, represents D~lu no constants arbitrary suppose that in the operation added (forour is to obtain a are objectnow and not the most general particular integral integral). Now since Dra~reax eax DrD-reax,it follows that
= =
Then
D~l
D~reax
=
a-reax.
=
Hence it is clear that Dneax aneax values of n positive or negative.
for all
integral
254
DIFFERENTIAL
EQUATIONS.
208. Let
(
=
pansion f(z) be any function of z capableof exof z, positive in integral or negative powers ^Arzr say, Ar beinga constant,independentof z).
Then
The
result of the
be obtained
Ex. 1.
f(D)eax operation may D by a. by replaci'ng
"
therefore
Obtain the value of
-
"
"
-e
^
Obviously by
the rule this is
"*
or
g.
E,
2,
Obtain
the value
of
By
the rule this is
e3a/'=-^-e
EXAMPLES.
1. Perform
the
indicated by operations
'
^
3.
Apply
Art. 208
to show
that
m^7 /(Z)2)sin mx /(Z)2)cos
=/(
=/(
"
"
m2)si m2)c
209. Next
Operation f(D)eaxX.
let y
=
eaxY,where
Fis
any
function
of
x.
PARTICULAR
INTEGRAL.
255
Then
we
since
Dreax
=
areax,
have
yn
=
by
Leibnitz's Theorem
-
1D Y+ nC2D2Y+...+Dn F+ n(71an edx(an the Binomial Theorem
F),
(Art,
which, by analogy with 206), may be written
Dneax Y=
n
F, + a)n eax(D
beinga positive integer.
Now let
we
X
may
so
that
write
Then
from
above
DneaxY=eax(D+a)nY
or
Dneax(D+ a)
therefore
in all
~
nX
=
eaxX,
and
D
cases
Hence
for DneaxX
values integral
=
of
n
or positive
negative
+ a)nX. eax(D shall have
=
210. As
in Art. 208
we
f(D)eaxX
eax may That is, the left of the by D + a.
side to transferred from the right D operator f(D) provided we replace
be
Ex. 2.
"
-
"
"
-
-
e2xsinx
=
e2x~
D
*
sin
x"
"
e~xsin x.
D2
"
4D + 4
256
DIFFERENTIAL
EQUATIONS.
EXAMPLES.
1. Perform
1
the
operations
o
1
'
1
h
(D
2. Show
-
If*
X
(D-I)*6
l
X'
D-l
that
211. We
and
Operation/(7"2)
have D2
sn wwu cos
=
(
-
m2) y
sin
mx,
cos
therefore
Hence,
a^
as
before,Arts.
j;/ r"9\
208
/v
"
and
o\
210, it will follow
sin
mx. cos
sin
J
n
D
^
)
'
mx cos
=
f( m2)
y
=
Ex.
eaxsin bx dx f
=
Z)-1eaxsin 6^7
6^ (Art.210) + a)~1sin eax(D
"
f"x/ X/^iSJlll
7)Wri
hv
f'Arf 91^ xxi li, Zi L 1
^
y
a
sin ".#
"
6
cos
bx
-_
62) ea*(a?+
tan-1- Y ^sin^.r
-
EXAMPLES.
1. Find
2.
of integrals si eaxcosbx, e^sin2^,e^sin3^, the operations Perform
by
this method
the
-sin
3. Obtain
2^,
-_
_"I L_cos,r,
sin 2^7. and
by
means
of the
cosine the results of the
values of the sine exponential mx. /(/")cos mx, /(Z")sin operations
PARTICULAR
INTEGRAL.
257
212. Let
Operation
us
next
consider the
operation
in of expansion F(z)is a function of z capable integral positive powers of z. then if no odd in powers of Z), Let F(D) be arranged the result may be written down occur by the powers rule of Art. 211. foregoing
where
Thus
""
^ sin S1T1 2#
=
sin %x"
"
"
~
"
sin 2#.
L-4
+ 16-64
51
occur we
But and
the
if both
as
even
and
:
"
odd
powers
the
even
may
proceed
follows powers
the odd
Group and together,
powers together then we may write
operation
sm mx
= ./T^ox
.
^
/TV,V
sin
mx
^
^
"
mx m2)sin
"
mx(
"
mx m2)cos
Upon
we
may
that in practice it will be seen examination write m2 for Z)2 immediatelyafter the step
"
writingimmediately
1
"
"\
"
f)~7
R
^
sin
mx"
E. I. C.
258
DIFFERENTIAL
EQUATIONS.
or'
r-y^
"
mi
"
"
SYT^ fjsrS
"
in/
)
"
"
J-'Xv
i
in
4^0 sin
i
.
ma;, etc.
,
Ex.
1.
Obtain
the value
of
"
"
"
-
="
sin 2#.
^
Thisis
sin D
2#, 2^,
"sin
-
J.O
or
^
Ex.
2.
cos
2^
"
^
"
sin 2^7.
Obtain
the value
of
^
-^e2*cos
^p. cos x
This
expression
=
e2*-^
" "
-
^
each [replacing
Z"2 by
"
1]
e2*
____
1
=
"
"(cos
4
x
"
sin
x).
EXAMPLES.
1. Perform
:-
the
operationsindicated
Z"3
in
the
pressions followingex-
D
-e*sin
x
+
260
DIFFERENTIAL
EQUATIONS.
EXAMPLES.
Perform the
operations
2)
CD+l)(Z"
2-
+
3.
-
J?
COsh
37
COS
07.
214. In
Cases
of
Failure.
methods of
to
applying the above Particular Integral, cases
met
of
are
failure illustrate
cases.
obtaining a frequently
course
with.
to
We be
propose
the
of
procedure
215.
The To Ex.
adopted
the
in such
1.
Solve
equation
(^L-y=ex.
dx
'
Complementary
obtain the
Function
is Ae*.
we
Particular
Integral
have
If
we
apply
Art.
208, the
result
becomes
i^i
We
may
or
""-
evade
this
and difficulty Art.
210
obtain
we
the
result
of
the
operation by applying
when
have
which
particular integral required. another of substituting method, however, Instead,
is the
let
us
examine
the
operation
=
"
-"
more
carefully.
of #,
we
Writing
x(\ +
h)
instead
have
263
CASES
OF
FAILURE.
Of may
this be
expressionthe portionLtex//ibecomes taken with the complementary function Aex
may
infinite,
; and
A
b
we arbitrary
regard A
A
+
-
as
a
new
constan arbitrary
/i
for
we
may
suppose
to contain
a
infinite negatively
por
to cancel
The The
the term I/A. xe* is the Particular term
desired. Integral vanish when h is decre;
remainingterms indefinitely.
The whole solution
contain h and is
Ex.2.
Solve
the
equation
^ Ct/X
is
The
complementary function
y
=
clearly
cos
A sin Zx + B
2#.
The and Art.
integralconsists particular
sin 2#. In this second
oo
,
of two
parts "i
we
"
e*
o
part, if
so
apply the
ri
211, we
now
get
2HL",i.e.
the
and
fail.
We This
consider
when limit,
h
=
Q, of
--
"
sin2.i'(l-
expression
1
=
_1
i_(i+A)a
1
9A
4
1
I
^X JT2^U
COS
"^IX "*" COS
^^ S^n
^^J?)
-
1 sin 2.2?
"
"
1
-x cos
o
fi
l
"
2.27 + powers
r of A
,
4
=
(a
term
which
may
x
-
be included
^
in the
complemvanish
v
function)
Thus the whole
c^s +
(terms which
solution
of the. differential
equation
JOS
g
"
"
260
3.
DIFFERENTIAL
EQUATIONS.
:.
Solve
'ere the
equation (D2+ 3D)(D -l)2y e* + complementary function
=
the
e~x + sin
x
4- x1.
is
plainly
parts,viz.,
=
consists integral particular
1
1
x_
of four
"?
*_ JL
'I?''
T
iy*6 ~(D-I*'
4~4
=
(a part going into
+"
ex +
the
complementary function)
vanish with
(termswhich
h)].
,^-PP,
10
1
-
3-Z)
sin x=
"/
6 + 2Z)
=
2(9
-
(3 sin
#
-
cos
^)/20.
jjinally
open
44
ie
the whole
=
solution
is
y
+ (A3 Al + A2e~3x
e2x
+
3 sin
~~
x
"
cos
x
.3?
.
5#2
,
44
+'+""l"
ILL
USTRA
TIVE
EXAMPLES.
263
Ex. The
4.
Solve is
the
equation"^-?/ ^sin^.
=
C.F.
To find the which
P.I.
we
have
."-a
is the coefficient of
i
in
1
"?". in
#*,
l.
" "*"
plX
rp
-4^-6/^Tr.
l
**
'
l
~^l-^D..\^
Thus and y
=
the
P.I.
is solution is
COSJP
_
8
3^ gin ^
8
the whole A
^inh ^
+
^2coshx
+
A3sm
x+A
x
4cos
x
+
c^s^' ^
-
sin
^p.
EXAMPLES.
1. Obtain
the Particular sina7'
indicated by Integrals
0)
7TTT
(5" /n (6)
_
w^ovn-Q-x*'# (sinh
+ sin
^?).
+ cosh
nT-T8*1111*'
(7) /na
^/
n
^o,(^
COS
-
6^).
COS
o
.
o
264
Solve the
DIFFERENTIAL
EQUATIONS.
2.
differential
equations
(3)
Cfc#
+y
=
(4) (D*-l)(D*-l)y=xe*. (6) (ZP-3D2-3D (7) OD3-%=#sin.". (9) (Z"2
(10) (^"+
(5) (Z)I)y
=
e-x
(8) (Z"2
216.
A
The
Operator
""-.
CvQC
transformation
which
renders
peculiarservice
in
reducing an
equation of
the class
where the
Av A2,
...,
are
coefficients
are
constants, to a form constants, arises from
x
=
in which
all
putting
et.
In
this
case
-TT
=
and e*,
therefore
at
x~ax
=
-^at
x-j-
It is obvious d -ji have
are
therefore Let
that D
the
operators
d for -j-.
and
we
dx
equivalent.
stand
Then
dx\
n nf\n
_
-*-
n
)X
_
Z.
" "
(x
\
__
11
-I-1 ]X
/
.
~
1
___
dxn
dx
dxn~l
EXAMPLES.
265
Now
putting11
in succession
2, 3, 4, ...,
we
have
etc.
,
Hence
generally
or
the order of the operations reversing D(D-l\D-Z)...(D=
Ex.
Solve
the differential
equation
Putting x
or
=
the equation becomes c?,
-
D(D -l)(D- 2)y+ 2D(D
%
+
3%
-
3#
=
(
(D
=
-
i.e.
giving y
Ae* + B
cos
-"
,
~+;rIog
EXAMPLES.
Solve the differential
1.
s
2
equations
dx
a?
2.
x-
+ --^
-^ + (^
=
+ x [log ^-]2
sin
log^
+
sin q
loga?.
3.
+
cfc1
4. "i
3^++2/=^
ote2 rfa?
'
dx?
5.
dx*
dx
CHAPTER
XVII.
ORTHOGONAL
TEAJECTOEIES.
MISCELLANEOUS
EQUATIONS.
ORTHOGONAL TRAJECTORY.
217. Cartesians. of a equationf(x,y, a) 0 is representative The to family of curves. problem we now propose is that of finding the equation of another investigate of which each each member cuts family of curves in of the former family at rightangles. And member such a problem as this it has been alreadypointedout The
=
that
it is necessary
to
treat
all members
of the
first
that so familycollectively, a ought not to appear in
It has
may
been
shown
be eliminated
constant particularizing the equation of the family. in Art. 17 1, that the quantitya between the equations
the
.
*dx
'dy dx
Let this eliminant
be
This
is the
differential
equation of
the
first
family.
268
DIFFERENTIAL
EQUATIONS.
Here
x+yJL=a,
cLx
and, eliminating #,
.v2+y2 2x( x -\-y" ),
=
Hence
the
new
differential
equationmust
be
or
^2+ 2^-^2
ay is
a
=
o,
...........................
(3)
become the be
homogeneous equation, and the variables by the assumptiony vx. separable as However, this being the same equation (2) with that x and y are interchanged, must its integral
which
=
ception ex-
another
set of
each circles,
of which
touches
the
#-axis
at
the
origin.
Ex.
2.
Find
the
of trajectory orthogonal
2
the
curves n\
i
--
A
being the parameter
of the
family.
and
A must
be eliminated
between
these
=
two
equations.
(2)gives
+ A) 0, + A)+yyl(a? x(b*
so
that
a2 + A
=
and Thus the differential
equationof
the
family is
(at-b^yy
or
x*-y*+xyyi-
=a2-52
................
(3)
ORTHOGONAL
TRAJECTORY.
269
Hence familv
changing y^
of
into
"
the
,
differential
equation
of the
#1 is trajectories
2
(4)
the
same
But
this
being
:
the
same
as
equation (3) must y2
i
a
have
primitive,viz.
^
*
n
_-.
~
i.e.a set of conic sections confocal Ex. 3. cardioides Here Find the
cos
with
the former
set.
of orthogonal trajectories
the
a.
family
of
r=a(l"
0) for
different
values
of
^
r^
dr
l~
= =
and, eliminating a,
Hence for the
sm
0
2 must
we trajectories familyof orthogonal
have
1 dr
"
n
or
log
r
"
2
log
cos
"
+
2t
constant,
or
r=b(l+cosO),
coaxial cardioides whose cusps
another
family of direction. opposite
point in
the
EXAMPLES.
1. Find
the
of the family of parabolas orthogonal trajectories of
a.
#2
=
4cM? for different values
2.
Show
that
the a2 b2
of orthogonal trajectories
the
m
family
of
.
similar
for different values of ellipses-04-^,=m2 the of orthogonal trajectories
a.
is s?
=Ayb
3. Find
r
=
the
equiangular spirals
confocal and coaxial
ae^cota' for different values of
4. Find
"
the
=1
of the orthogonal trajectories
-f cos
parabolas
9 for different values
of
a.
270
5. Show
DIFFERENTIAL
EQUATIONS.
of
curves
that
the families
are
orthogonal.
6. Show
r
that
=
the
"
curves cos
sin2a
a(cos 0
that
a)
=
and
r
sinh2/? a(coshft
=
"
cos
0)
are
orthogonal.
7. Show
if f(x+iy)
u
+
iv
the
curves
form
orthogonalsystems.
that for any cosh
x x
8. Prove
constant
cosec
value
p cot y
x
=
of /z the
constant constant
family of
curves
y
"
cut at
family /z coth rightangles.
the
"
cosech
cos
y
=
[LONDON, 1890.]
SOME
IMPORTANT
DYNAMICAL
EQUATIONS.
220.
The
equation
of
a
general form of the equation of motion the action of a central force. under particle
is the
Multiplying by
2-^and
dO
we integrating
have
which
we
may
write
as
and
the solution
is therefore
effected.
221.
Equations
of the form
SOME
SPECIAL
FORMS.
271
with
stant con-
have The
alreadybeen
coefficients. solution may
sin
discussed however
as
beinglinear
be conducted
to be
an
thus
:
"
Multiplyby
factor.
n9, which
will be found
integrating
Integrating,
sin
nO^. d6
-
nu
cos
nO=
sin n"dff f*f(ff)
+ A.
Jo
an
nO is cos Similarly, first integral is
cos
factor integrating
and
the
ing correspond-
n(" +
d\j du
nu
sin nO=
f'f(ff) nO'dO'+B.
cos
J
o
-^L Eliminating
nu
=
ef(0') sin n(0
0
-
O')d0f + Bsmn6-A
cos
n6.
222. The
mass
equationof
some
motion form
of
a
body
of
changing
often takes
such
as
d! dt
^
and
for this
will equation"f"(x)-rr
be found
to be
an
factor. integrating
leads at
once
to
j^x)"l"(x)dxA, i{""(^' J2=
+
1
J
^Xto
and
the variables
are
separated.
272
DIFFERENTIAL
EQUATIONS.
FURTHER 223.
one or
ILLUSTRATIVE
EXAMPLES.
Many equations may
be solved forms
other of the known artifices. special
Ex. Let
1.
by reducingto discussed by already
^=f(ax+by).
'
ax
ax+by=z.
=
Then
dx
dx
Thus
dx
dz
or
x+A=l
d*..
J
a
+
bf(z)
Ex.2. dx\ Put
rpi
y+a
dx xy=z.
Then
y+^^=-y-,
.
dy
dz dx
dx
dz
Z or
=
1
, X-j- +-5-
dx
dz
dx which is of Clairaut's
form, and
the
is completeprimitive
.
Ex.3. Let
Solve
e^
-
dx)
6^ =
\dx
77,
ex
=
s-
Then,
since this
equation may
dx
he
arrangedas \e*dx
ILL
USTRA
Tl VE
SOL
UTIONS.
273
we
may
write it as
?7
which written
being of
Clairaut's form
the
completeprimitive may
be
or
Ex.
4.
--
in occurring (an. equation Put Then the
Solid
=
Geometry).
,v=*Js equationbecomes
and
y
"Jt.
ds
,
giving
t=
ds
as
which
is of Clairaut's form
and t-sG
has the BC
~
completeprimitive
1+2(7'
and
solution singular
the four
lines straight
9"J-Jy
Ex. 5, Solve the
equation
dx
E. I. C.
S
274
Let
DIFFERENTIAL
the transformation be such
EQUATIONS.
that
then
x
is known
by
direct
as integration
a
function
of
t.
dy
Now dx
d^dL_
*
'
and dx*
Thus and the
f (^ax^yJ^axd4 }dx* dt* dx dt
.
,
givenequationthus
reduces
to
whose and is
solution is y=A sin qt + B when the value of t in terms have dx
cos
"^,
x
of
is
the substituted,
solution
complete. we [Ifa be positive
1
"
,.
"
-[=.
""~j= sinh^Wa)
If
a
=
t.
be
we negative
have dx
1
,=dt"
V-a
are
differential Solve the simultaneous Ex. 6. linear with constant coefficients)
equations(which
276
whence
DIFFERENTIAL
EQUATIONS.
D2 + 9 and
these in turn by operating upon eliminate y and obtain we subtracting,
+ 16)(D2 + 9)+ [(D2
or
by
3D
and
15
Z)2" 0,
= =
+ 40Z)2 + (Z"4
1 44"
0,
i.e. whence
x
=
+ 36)^=0, (D*+ 4)(D'2
A sin
~2t +
B
cos
2" + C sin 6" + D
cos
6*.
the Differentiating the second
to eliminate
three times first equation and subtracting have differential coefficients of y, we
dt whence viz.
:
"
*"
we
obtain
the
value
of y
without
any
new
constants,
y=-%B
sin 2t + 2 A
cos
2t + i"D
sin 6"
-
^-
EXAMPLES.
Solve
i.
the
equations
2. 2^-(i-*)y"=**. 'cte
3.
4.
5.
(1-
.
2 8. Obtain
:
"
2
cosy the
the
integralsof
following differential
tions equa-
+
9y
-
25
cos
^
[I.C. S, 1804.]
EXAMPLES.
277
9.
Integrate
the
simultaneous
system
4=0.
_
10. inclination
to
Find
the of the
form
of
current
the
curve
in
to
which the the which
the #-axis
tangent
is
of
the
tangent
of for
proportional
the
11.
product
Find cube the of
of form the
the
coordinates of the of
point.
the of the
curvature
curve
varies
to
as
the
cosine
the
inclination
tangent
the
12.
Show of
that
curvature
in
on
the the
curve
for is
which of
constant
the
projection
of
the
radius
7/-axis
length
(l)Soclogtan(?+|),
(2)
y
oc
log
sec
~.
ANSWEES.
CHAPTER
I.
PAGE
12.
1.
Area
=
e6-ea.
3.
Area=ia2tan
0,
4.
Vol.^.
5
2.
Vol.
=-(e*b-e2n).
Vol.
=
-a3tan2"9.
5.
Vol.=f7ra3.
6.
(a)
Vol.
=|
Vol.
=
1
TT
VoL
1
=
*
(8)
-
JL25
u
Vol.
=JL
t)
7,
"7Tfia3.
8.
Mass
of
half
the
spheroid
=
J?r/xa262.
ANSWERS.
279
CHAPTER PAGE 17.
II.
a
2.
^Y.
6. 1.
10.
+
6 -sin a). (sin
3.
?^1.
Ioge|.
7.
x/2-1.
4.
8.
|.
PAGE
23.
"2
X
"'
*'
ft C"
X
r!00
Ht;
r!000
"
r!001
loo' Tooo' Tool'
_c.o
10'
_^
100'
_^
98
PAGE
25.
"
, 2J
2.
a
logx,
~j
a
log^ +#,
PAGE
26.
2.
3.
logtan"1^,logsin'1^, log(log^).
280
INTEGRAL
CALCULUS.
PAGE
28.
9
"
log2J
""
4+
aT
+
4
log3'
_?
log 6
+I
logo2'
4.
log tan
^,
log sin
^
-
cosec
^.
5.
sin-1*,1 tan-,
l86^! an-^,
7.
8.
-log^
+
sin x\ e*), log(log
CHAPTER PAGE
1. 32.
III.
sine*, sin#n, sin(log^).
3.
asin^+-tan-1^4,
4
-a
cose*
+6
log cosh
8.
#.
4.
J_ tan-x-JL-.
V2
_-,
6.
~.
sin-V^-
2
^2
"
3
.
L
V*
'
v
PAGE
/-
41.
g
-1?,
-
"
-H-.-fJain-1*,
2.
cosh^+l),
si
ANSWERS.
281
3.
-Vl=F, x/^,
s
4.
+1)4, i(^2
6.
-^2,| sinh-1^ I siii-1^-2\/r^-iWl
+
2\/l +
#2 +
7.
xyJ\^3?,4 cosh-12
+
^"?x/5?^
2
8.
^logtan^,
-logtana"r+", ^logtan(-+.A
CL
2
\4
/
\
f
x
),J logtan
-
13.
Iog[log{log(loga7)}], log(log^), log{log(log,^)},
CHAPTER
PAGE
IV. 47.
^7
sinh
x
"
cosh #,
x (2+ #2)sinh
"
Zx cosh 2^
,
x.
x
sm
2"r
CQS
in 3a? + 9 sin
3. 2a? J(sin 2^
+ cos a?)
3^ + 27
cos^].
-
cos
2^),
sin 4.f
~""
sin 6^7
~""^~~
_
/cos 2^7
cos
~~
4^
_
cos
~~
6^7
282
INTEGRAL
CALCULUS.
5.
--^sin^-tan-^),
2
v
^sin^
-
tan-a4).
of n, -p-q-r.
5
6.
^2(a2
+
^2)~sin^-tan-1-,
\
ct /
for the values q + r-p,
p + q-r, 7T2
I.
r+p-q,
7T,
_,
A 7T2-4.
2
9.
sin-1*?+
339
x/f^;2
-
(1 ^2).
-
PAGE
51. + 2)cosh 3(^2 x, 2^72 + 24)sinh x. + 1 5(^4
-
+ (rf
20^"3+ 1 20#)cosh x
-
6)sin x,
_
84\V
-
2/
-
\ 2
4
2^ J{2(2^3 3^)sin
-
(2^4
-
6^2 +
3)cos
5?r4+ 607T2
-
240, 265e-720.
52.
PAGE
1.
(a) (m2+ (6)
4
/
where #=sin#. l)~^rnecos(^-cot-1m),
+
gsmg
\
cos^-sn-g,
3 3
/
where ^74
"
" -"
A-
=
Si .tT3 3^7
"
(c)^
\
tan
^
+
log cos
"
x.
(e)
/
\
1
tan
,
_i
lx
"
"
"
(d) ^tan"1^ (a) x
"
Jlog(l+^2).
(/)
x
sec"1^
-
cosli"1^.
\/l" ^sin"1^.
+
+ cos0)-sin"9-logtan (b) 6"(sec(9
Y
where
^
=
sin(9.
(c)
2
where cos (c?) (sin"/" "^" "/"),
"
"
^=
284
INTEGRAL
CALCULUS.
CHAPTEK PAGE
58.
V.
2.
"
.
3. 4. 5.
+ 4# \ log(^'2
+
+ 2). 5) taii"1^
-
-log(3-.r). #-2log(#2 +
-
2.r +
2)+ 3tan-1(."
11
6. 2#
+ 6^7 + 10)+ f log(#2
tan-1^ H- 3).
PAGE
62.
^ (iii.)
-
_
ct
"
o
x (ix.)
+
7 a^T
-
giZ^
+etc. log(^ q.)
-
-
e
1
(^Tiy
+ 8
+
(^-l)2 8(^-l)
16
j B
27
I_L.-?-JL_. l0a^"^ + g^-l+
