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CHAPTER XV
MORE DIFFICULT INTEGRALS:
INTEGRATION BY PARTS

Differentiation can be applied to any continuous function of x by the application of simple and straightforward principles. The inverse process of integration is at the best a tentative process and depends very largely on whether the function to be integrated can be recognized as the derivative of another function. It may happen that, although the function as it stands is not familiar as the differcoefficient of another function of the variable, it can be so transformed as to be immediately integrable. A simple example has been given in the previous chapter, where, knowing the

standard form

)

dx,

we can derive at once the integral

(a2

+ x

)

with respect to

The integration of more complicated functions is largely a matter

of practice. There are, however, certain standard methods of attack, which, although they may not invariably produce the required result, can often be applied with success to the solution of the problems of integration.

The method of substitution. Let us consider the problem that is denoted by f dx. In the

first place, the

shows that the independent variable is

Secondly,

is a function of

such that if we know, or can find,

another function of

whose derivative with regard to

then z is the required value.

Put shortly, if

then

dx = z.

ⁿ⁺¹

dx =

(

ⁿ

-I-

x"',

dz dv

^=y,

A familiar example is

236    INTEGRAL CALCULUS
d    ^I+~~
or    _dxn+Ix=x

so that if y = xⁿ, then
f
ydx=_n+xⁿ⁺ⁱ.

Suppose that y is a more complicated function of x, say (a + bx)ⁿ. We do not immediately recognize j y dx, and it is necessary to proceed further.

Let    z=a+bx.
Then    dx = b,
and    dy _ dy dx — dy I dz dx dz dx b^.

If therefore we replace the independent variable x by the new variable z, any differentiation with respect to x becomes a differwith respect to z by the simple process of dividing by the constant b.

The integral I y dx may therefore be written as fy [b dz] , and, since y = (a + bx)ⁿ = zⁿ,

J y dx = J (a + bx)' dx = Jz^t'dx

f - zn [dz]

_-_Jbzⁿdz

₂

n+1

+ I

b n + i (a + bx)

^n+I

This is an example of the method of substitution.

Again, to evaluate

y dx

where

is the function

sin

+ cos x

let

cos x.

THE METHOD OF SUBSTITUTION ²37

dz
dx = — sin x,

dy dydxdy r_ I

dz dx dz dx I sin x
Therefore, as above, J y dx becomes
I
sin x
y dz;
f    sin x    sin x    _ I
)I+cosxdx _ ^—~I+cosx    sinxl dz
    ^I
— [ I+cosx dz
I
I +_z^dz= — log(I+z)
= — log (I + cos x).

When applying the method of substitution it is customary to shorten the initial process. If the substitution is z = cos x, we write, instead of

— sin x,
the relation dz = — sin x dx,
so that in the integral we may immediately replace dx by
I
sin x_
dz.
In general, if the substitution is f (x) = (y), instead of writing f^' (x) = (y) _dxwe write immediately f^' (x) dx = (y) dy. It

will be observed that the process is to differentiate each function with regard to the variable in which it is expressed, x, y or z for example, and then to multiply by dx, dy or dz as the case may be. The expressions f^' (x) dx, (y) dy, ... are termed differentials of

f (x), (y), ... respectively.
Then
dz
x

238 INTEGRAL CALCULUS

For the purpose of integration this procedure should be looked upon simply as a convenient means of passing from one variable to another, and not necessarily as a splitting up of the dx into two parts dz and dx.
We may consider the problem from another point of view. The definition of a definite integral is the limit of hEO (x) between x = a and x = b — h, as h tends to zero. Since h denotes a small increase in the

value of the variable, we may equally well write the integral as x=b–ax
Lt E 0 (x) .Ox = Lt [0 (a) Ox + 0 (a + Ax) Ax +
ox--> 0 x–a
+ , (b — Ox) Ax],
b

or' _cc (x) dx, where the Ax is the small increase in the value of x

.a
which tends to zero.
The method of substitution is then as follows. If, as in the second example above,
z = cos x,
then    Oz = cos (x + Ox) — cos x
_ — 2 sin (x + 20x) sin
Now    Lt — sin (x + 20x) = — sin x,
0
and    Lt sin 1-,,x — _rAx—)-o
2~^x
Lt 0 (x) Oz = Lt {— 2 sin (x + 2Ax) sin 2Ax. _qc (x)} Az-÷o    .1x, o
— 2 sin x Lt 0 (x) 2Jx
_ x—> 0

= — sin x Lt 0 (x) :,x.
.ix -).0

Lt 0 (x) Ox = [—   .¹Lt 0 (x) Oz, sin x o.~0
which gives the same result as that found above,
3. If y = f (x), then 0 (y) = 0 { f (x)}, and if f^'(x) denote, as usual, the differential coefficient of f (x) with respect to x, then
(y)dx=~{f(x)}f'(x)
i.e.    0 (y) dy = 0 { f (x)} f' (x) dx,
so that    f (y) dy =    { f (x)} f' (x) dx.

THE METHOD OF SUBSTITUTION    ²39

Now on the right-hand side the integrand is the product of two functions, qS { f (x)} and f' (x). If we associate f^' (x) with dx we have (i) a function of x, namely {f (x)}, and (ii) f^' (x) dx, which is a differential. It is evident therefore that if it is required to integrate an expression consisting of the product of two functions, one the differential of a known function f (x) and the other a function of f (x), the substitution f (x) = y will reduce the integral to the

simpler form J qS (y) dy.

It must be noted, however, that if the product is to be simplified in this manner, both conditions must be satisfied. The form of substitution is frequently determined by the recognition that the expression to be integrated contains f^' (x) dx, the differential of f (x).

For example, consider the integrals
x _/Idx_x2'Jf _/1³dx_x2'v I      x2 dx.
These may be^e written    V
2
^I--- xdx; J    x      _2xdx;f~/I2 ^x2xdx,
x²1/I+x    x

and it is evident that each of the expressions consists of x dx (the differential of 2x²) and a function of 4x2. The substitution of y for x²will therefore simplify the process of integration for all three examples. It will be observed that this substitution is suggested partly, if not wholly, by the presence of x dx.

Again, consider
dx    x²dx    ^fV I + x2 dx.
VI + x^2'1V1 + x2'    x²-¹It will not help to write these in the form J (x²) dx, as the
required xdx is absent. The substitution y = x²will not therefore simplify the integrals.
4. Further examples of substitution.
It does not follow that any integral can be evaluated by a simple substitution nor indeed that the simplest substitution is the best.

240    INTEGRAL CALCULUS
The examples given below are merely indicative of the methods to be employed: they are not necessarily of universal application. Example 1.
f x² (a + bx)' dx.    (n* I.)
The substitution is    z = a + bx.
We have    dz = bdx and x = ^{z a}

₁₌( (z — al^e I I dz I ` b ) zⁿb

^"z²— 2az + a² ^II - --z_n dz
bs
= Ib3 (z2-n — 2az^'-n + a² z^-n) dz
= b³L rz2-n dz — 2a I z'-ⁿ dz + a² f z-ⁿ dz]
I I z.3-n — 2a z^2-n+ a² z'^-n

-b³ 3—n 2—n I—n

n is

a positive integer greater than

this result is more conveniently written as

-b³ L(n — 3) z^n-3(n — 2) zⁿ-² + (n — ^a) z^n_i] wherez==a+bx. The above method can be applied to any function of the form

(a + b x

where

is a positive integer. It is easily seen that the

)

ⁿ

integral will be _b,m}₁ I ^'(z _ zna dz. By expanding the expression in brackets by the binomial theorem and integrating each term separately, the result follows.

Example

(2xdx

I+

_x2

The facts that 2xdx is the differential of x² and that the remainder of the integrand is a function of x² suggest the substitution x²= y, or, better, I + x^'- = y.

ILLUSTRATIVE EXAMPLES 241
Put, therefore, z = 1 + x², so that dz = 2x dx. Then
(_Id₂ j₁ dz_x2 j z^z=logz=log + x²).

This is an example of a general proposition. Where the numerator of a fraction is the derivative of the denominator, the required integral is the logarithm of the denominator.

In other words:
dy
For example,
^dx_f    dx=logy.
d (1 + cos x)
r +^scos x d
in x
1 +^dco
x
s _xdx = log (1 + cos x).
(Cf. para. 2 above.) d sin x
Again, f cot xdx =1' cos x d x =    ^dx dx =log sin x.
sin x    sin x
Example 3.

Now
dB
I sin B^'

sin 0 =
2tan,B
I + tan² ,B
_I_ I' 1 + tang _zB dB.
2 tan 2B

(Chap. I, para. 16.)

Put t = tan 10,
dt = 1 sec^t 4Bd0 = z (1 + tan² 40) dB.

^dB= + tang _2B

I + tan² 'B    2dt    _ f    dt
_J'dt
2 tan 10    + tan^g 20    tan 2B    t = log t
= log tan 10.
F    16
2dt

242    INTEGRAL CALCULUS
Corollary:
r    de    dO ,Icos0^—f sin( it—0)
0)
=— log tan -_,,^I,-_,(ill- - 0) _ — log tan (err — 20)
log tan Orr — tan 20 I + tan Orr tan 1-0
_ — log I — tang    (since tan Orr = I) 1+tan₁0
I + tan
_
log I — tan ₁10
Example 4.

dx ~/_x2 + _a2

Method (i). Put x = a tan a, so that dx = a sec² ada.
    I=         asec' ada
I
J 1x² + a²
f    I
_I     asec²ada 1'a² tan² a + a²
I
    asec² ada a '/tan^s a + I
= I sec ada    (since tan² a + I = sec² a)
_ _f( da
= Jcosa
log I + tan la
    —    (from Example ₃above), — tan
tan a = x/a.
Method (ii). Put Vx² + a² = z — x.
x²+a²⁼z²-2zx+x²,
a^t = z^t— 2zx.
o=2zdz—2zdx—2xdz.
— x

dx = " dz. z

where
Then or

ILLUSTRATIVE EXAMPLES ²43
dx — _iI z—x _ ~dz :. ^I=J1/x²+a2 z—x z dz z
= log z

= log (x + 1^/x²+ a²).

This result is the same as that produced by the first method. For, if x/a = tan a,

log (x + 1^/x²+ a²) = log (a tan a + 1/a² tan^g a + _a2) = log a (tan a + sec a)
=loga+log{sing+ I 1 Cosa cosa
= log a + log sin a + I cos a
I —tan2a^'
which is the solution given by Method (i), since log a is a constant, and the result of differentiating

to ^I+tan,a+loa
^gI —tanla ^g

is the same as that of differentiating

I + tan a
^logr — tan zQ + any arbitrary constant.

Method (iii). Put x = I/y in order to obtain an odd power of y out-side the square root.
Then dx = — (r/y²) dy,

x² + a² = ^I/y2 + a²= (^I+ ^a2Y²)/Y^2.dx r y r
JVVxx2+a²⁼J1/1 +a2y2 `- y2! Y
_If —dy
–,/ y ~I +. a2y2

2 tan 2a
I + tang-a + I
=loga+log_l    2~
I—tan za
I+tan²2a
= log a + log (I    + tan₂2a)²
I — tang
= loga + log i + tan 2a
i6-2

²44    INTEGRAL CALCULUS To eliminate the square root, put I + a²y²= z².

	2a²ydy = zzdz,
and	I = ^{f _}z dz _ f — dz
and	I a²y--z — 1 z²— 1'
since	^(z2— I)
	y- _ - _a2

On integrating and substituting, first for z in terms of y, and then for y in terms of x, we have
I=2log
1x²+a²—x
which is easily seen to be log (1^/x² + a²+ x) — log a on multiplying numerator and denominator of the fraction by 1/x² + a² + x. Corollaries :
(i) J      dx
a2 log (x + 1/x² — a²)
^J1/(^x =+(^x — b) .I1~x2 - (a dx
b₎ x + ab (ii)
^—J1 {x—i(a+b)}2—{ (a — b)}²
= log {x — (a + b) + 1/(x — a) (x — b)} =log {(x—a)+(x—b)+217(x—a)(x—b)} =log(1^/x—a+lax—b)²—log2 =2log(1/x—a+l'x—b),
disregarding the constant of integration.
Example 5.
_I    +     x3 is not recognizable as the derivative of another function of x. We proceed therefore to express it in partial fractions.
I    I I    I    2— x

I+ x³31+x⁺31-x+x^2.

dx
I+x—log(I+x) at once.
2 x 2 dx needs further investigation. r— x+x
yx²+a²+x

dx
I

ti
or

ILLUSTRATIVE EXAMPLES ²45 Now if the fraction were of the form
df (x)

f (x) = I - x + x²and d d-x) = - I + 2X,

which is not the numerator in the given integral. If, however, we express the numerator thus :
2-x⁼ - (- I +2x)+
the integral becomes
    (- I ( - I + zx t d x + [^{3      I}dx.
2 {l1 -x+x²J    2 I -x+x²
The first of these is — i log (I — x + x²), and the second may be written in the form
f3    ^I
2 (x — 1)² + ('/3/2)2 dx. This is of the form
1 ³dx,
2 x²+ a²the integral of which is
x 3 I - tan^-1

2 la    a)
r
²(^x— 1)²+ (V3/2)2 dx = 2 V3/2 tan-1 A/₃12
.~ tan^l 2_x —I "V3    V3
The complete integral is, therefore,
f dx        - dx    I I - I + 2x    I 3      I    dx I+x³ 3.~^I+^x 3^f 21-x+x2dx+3 ¹²(^x_1)²+(/3/2)2 = - log (I +x)—₆log(I —x+x²)+ I-tan-^12x-^I
3    1/3
Example 6.
- / xn      dx, where n is an integer.
Vx²+a²
From a consideration of the illustrative examples in para. 3 (^p.²39),
dx
7 (^x) '

the integral would be log f (x). Here

246 INTEGRAL CALCULUS
it will be seen that if n is a positive or negative odd integer, the integral can be simplified at once.
Let n be 2m + I. Then the integral becomes

J 1x22+ a2 xdx,

and consists of the differential of 2x², namely xdx, and a function of 2x². The substitution 2x² = y, or, better, x² + a² = y, will therefore simplify the integral.

When n is even it will be found that, in expressions containing Vx² + a², the substitution x = 1/y has the effect of changing the index of the term outside the radical from an even number to an odd number.

Let n be 2m. The integral becomes
2m
ix 1x² + a2 dx. Put x = I/y, so that dx = (— 1/y²) dy.
I I

I ^=j—y2m 1,(I/y2) + a2 ' ^dy

I
and the index of ^y2m+tbeing an odd integer, the substitution I + _a2_y2 = _z2 will now be effective.
Corollary. Since
=1— y2m+I I + a2y2 dy, (bx + c = a    b
ax² +    x²+ - x+^c-\
a    a
b .= a [(x + 2a~2 + \ (c ^b2a 4a² )J '
which is of the form a (x² + k²), we may integrate by the above methods functions of the form
xⁿVax² + bx + c

5. Forms of integral which can be evaluated by the application of general methods are those involving irrational expressions of a simple linear or quadratic type.

Type (i).

dx
J(x+a)Vx+b^.

SIMPLE IRRATIONAL EXPRESSIONS    247 Since the only consideration is the elimination of the radical, put Vx+b=z, sothat x=z²—b.

Then    dx = 2zdz,
i.e.    dx    — dx = dz,
2 VX + b 2z

and the integral becomes

( 2dz — ( 2dz
x + a , z²+a—b'
which is immediately integrable.
Corollary. The form

(x + a) dx
(x + c) x + b

is evaluated by writing the integral as

L 1"x¹+ b + (x + ^ac) Vx + bJ dx.

The first integral is a standard form and the second is of Type (i) above.
Type (ii).
dx
(c positive).

₍x k)Va+2bx+cx²

Several methods are available here, the procedure depending upon the particular substitution adopted. We may consider either the quadratic function or the linear function as suitable for the substitution, but in neither case is the process immediately obvious.

(a) Let    Va + 2bx + cx² = z — x C

Then    a + 2bx + cx² = z²— 2zx V + cx²,

or    a+2bx=z²—2zxV

2bdx = 2zdz — 2 Vc (zdx + xdz), z—x V
and    dx = b + z ~/c dz.

2₄8    INTEGRAL CALCULUS
The integral is therefore

f    I    z — ^x\cdz J (x — k) (z — x Vc)b+zVc
dz
(b + z VC) (x — k)^'

and, since    a + 2bx = z² — 2ZX Vc,
z²— a
x=
2 (z VC + b)
so that the integral takes the form
J_(b + z V c-) [z2_a 2 (z V
C + b) — ki
    2dz
z²—a— 2k (zV^'c+b)^' which is of the simple rational form

dz Jz²+Cz+D

(b) Let    x — k=z, or x=k+z;
I
dx = —
_z2—dz.
    dx
J (x — k) Va + 2bx + cx²
J — I    I    dz
z2    {a+2b(k+z) +c(k+z)²¹
    = — ^Iz    ^I    dz_,
2
z i vAz²+Bz+C
7
where Az² + Bz + C = (a + 2bk + ck²) z² + 2k (b + c) z + c.
    II    dz
I _ —J VAz²+Bz+C^' which is immediately integrable.

then

INTEGRATION BY PARTS ²49

For further information on the subject of these integrals the student is advised to read Williamson, Integral Calculus, chapter Iv. Another general method will be found in Henry, Calculus and Probability, where the required substitution is obtained by putting one of the constituent functions of the expression equal to y^r. The index r is then determined so that the integral can be evaluated by known processes.

6. Integration by parts.
From the relation Auxvx = u_xr1Av_x + v,Aux a formula for the finite integration Eu_xv_x was obtained. Similarly, since

d dx^(uv)=udx^+vdx,

we may derive an expression for the integration of the product of two functions of x.
For, integrating both sides, we have

uv=_Judxdx+iv^dx
d
dx,
or ludxdx=uv — ivdxdx.
Replace u by U and let dx = V, so that v = ( Vdx.
Then f UVdx = U f Vdx — f (dx f Vdx) dx.
If therefore _[dxJVdxl is integrable we can at once find the value of f UVdx.
In words the formula may be written thus :

The integral of the product of two functions of x = (the first function x integral of the second) — the integral of (the differential coefficient of the first x integral of the second).

A few simple examples will show the application of this process
Example 7.

250 INTEGRAL CALCULUS

The point to consider at the outset is which of the two functions should be taken as " the first function " and which " the second.^" Take x as the first function; we are to differentiate the first function and the differentiation of x will produce a constant.

fxe^xdx=x le^xdx— I (dx I e^xdx) dx = xe^s — J e^xdx = xe^x — e^x. ix log xdx.

We must choose log x as the function to be differentiated, for if we take x as the first function we shall have to find the integral of log x—which is not apparent.

Example 8.
z

=₂ log x — 2 dx

x² x² = 2 log x — 4 .

7. The method of integration by parts is useful even where we have to integrate a single function of x. We may treat f (x) as the product of two functions, one function being f (x) and the other unity.
Example 9.

~tan-¹ xdx.
Let the first function be tan-¹ x and the second function 1. Then

f tan-¹ xdx = f tan-¹ x. xdx
= tan-¹ xi Idx — I {d_x (tan-¹ x) f I dx} dx

^JI
= tan-¹ x. x — ₁+ _x2xdx

= x tan-' x — I log (x + x²).
J

x log xdx = log x I xdx — I {d dg x ( f xdx )} dx

^l_x2
x
-₁x2
`    /
=logx₂—I- dx
x    ^JJ

INTEGRATION BY PARTS    251
Example 10.

f \/x^z +a²dx. As above

f 1^Vx² + a²dx = f Vx² + a². idx
= "Vx² + a² fI dx — {_dx'Vx² + a² f1 dxI dx
x
= Vx2 + a2.x – _,v_- / + _a2xdx
x2      dx
= 'Vx² + a². x – ~V x² + a2
x² + a²    a²
Vx2 + a². x –    dx
j₍    ---- -
^-/x²+ a2 Vx2 V+a^2J
a²dx
=1x²+a².x- gVx2+a2dx+ ⁱ_x²+a² dx 21Vx2+a2dx ^=xVx2+a2+a2~./_x2_+a2
⁼x^VVx²+a²+a²log(x+\Vx²+a²). v x² + a²dx = {x Vx² + a²+ a²log (x + Vx2 + a²)}.
This example is instructive in that the process of integration by parts does not immediately give the required result. When we have performed

the necessary operations we are left with I \/x2 + a²dx on both sides

of the identity, and we have to clear the right-hand side of this integral before the answer is obtained.
A somewhat similar process is necessary in the evaluation of the following important integral involving trigonometrical functions.
Example 11.
f e^x sin x dx.
Here it is immaterial which function is chosen as the first function. Take e^xas the first function: then

252 INTEGRAL CALCULUS
f e^x sin xdx = e^x[sin xdx — r{dx-x [
sin xdx} dx

= — e^xcos x + [ex cos xdx.

This does not immediately give the required result. If, however, we consider fe^z cos xdx and integrate by parts, we obtain a further equality which enables the integral to be evaluated.

f e^xcos xdx = e^xfcos xdx — r{dx f cos xdxj dx

= e^xsin x — f e^x^Jsin xdx.

Let (ex sin xdx = S,

'e^x cos xdx = C. S= — e^xcos x + C, C = e^xsin x — S,
S — C = — e^xcosx, S + C = e^xsin x.
fez sin xdx = S = 2ex (sin x — cos x),

(ex cos xdx = C = le (sin x + cos x).
8. Reduction formulae.

It has been shown in the preceding paragraph that, in certain instances, integration by parts may not produce the required result immediately ; another stage must be reached before the integration can be effected. It is often possible, however, to relate an integral to one or more integrals of similar form, so that by proceeding successively the original integral can eventually be obtained. If, for example, we are required to integrate u_n, where u,, is a function of x involving xⁿand lower powers (n being a positive integer), it may be possible to relate the integral of the function to the inof u_n_₁. The formula connecting these integrals is called a "reduction formula.^" Reduction formulae are of importance in the integral calculus, and their use often leads to the evaluation of

and
Then and or
Hence and

REDUCTION FORMULAE ²53 integrals which could not otherwise be obtained. For the purpose of illustration it is unnecessary to give more than a few elementary
examples of the application of these formulae to indefinite integrals. It will be seen later (Chapter xvi) that the process may be adopted to greater advantage in considering problems involving definite integrals.

Example 12.

f e^xxⁿdx.
u_n = re^xxⁿdx = xⁿe^x — fnxⁿ-¹e^xdx

^J= xⁿe^x
— — ^nun_l^.
Similarly u_{n_1}= _xn-l_ex — (n — i) ^un_2,
and so on.
The integral can therefore be made to depend upon the value of u₉,
i.e. on /e^xdx.

Thus f e^xx³dx = x³e^x — 3 Je^xx²dx, fe^xx²dx = x²e — 2 /e^xxdx, (e^xxdx = xe^x — f e^xdx

^J=xe^x—e^x.
fe^xx3dx = x²e^x — 3 [x²e^x - 2 (xe^x — ^ex)] =e^x(x³—3x²+6x— 6). Example 13.

ftanⁿ Ode,

where n is odd.
tanⁿ 6 = tanⁿ-²6 tan²O = tanⁿ-²0 (sec²O — I). /tanⁿ Ode = f tanⁿ-²9 sec²OdO — f tanⁿ-² OdO.
To evaluate        l J tanⁿ-²0 sec²OdO,
put    tan 0 = z.

254    INTEGRAL CALCULUS
Then    sec² OdO = dz,
    x4-1    _tn 1 0
and    rtanⁿ-² 0 sec² 8d0 = Jzⁿ-²dz = n — I    o — i
I
.', u_n= -- tan^n-1 0 — u_n_₂n —
    tanⁿ-¹ 0 —    tanⁿ-³ B +    tanⁿ-⁵ 0 — ... ,
n — I    n -3    n—5

where the last term is (— 1)^m I tan Ode, since n is odd (= 2m + 1), or
(— 1)m+l log cos B.

Many reduction formulae result from the differentiation of simple functions of the variable. It will be sufficient to give one example of the process.
Example 14.
Find a reduction formula for the evaluation of the integral

xⁿ
dx,
1^/I + x²

where n is a positive integer. We have identically
d xn-1,V1 + x²= (n I) x^n-21^/I + x² + — ^x- - - x^n-1dx ₁/_I + x2

(n — I) x^n-2(1 + x²) + xⁿ
1^/I + x²

(n — I)x^n-2+(n — I)xⁿ+xⁿ
V I + x²

xn-2    n
    (n — 1) 1/    + nx
I x²1^/1+x²
By integration
n 2    n
x^n-11^/I + x²= n _ 1) r x - x + n" x dx
    (    !1^/I+x²    . VI+x²~

I
i.e.    u_n= - x^n-11^/I + x²— n

n ^nn-2'

n — I

ILLUSTRATIVE EXAMPLES ²55 By continuing the process we arrive at the forms

f Vl^dx
x2 (n even), or r-I ^d+ _x2 (n odd), which are immediately integrable. ^J

9. It is evident from the explanations above that the evaluation of an integral may depend on one or more of a number of different artifices. For complicated functions it may be necessary to resort to several alternatives before the solution can be found. Indeed, in many instances the integral may be no known function.
The following example is of a different type from those hitherto examined, and is given in order to show that certain obvious submust be rejected as being unsuitable for the evaluation of the integral.

Example 15.
dy
^f(I — y³)^I

Let I = ^dy (1 — y3)- .

(i)Put I — y³ = z, so that y = (I — z)^I. Then— 33'² dy = dz.

_I=

I I

dz= —

J 3.Y

3 J z

(1 –

which has produced a more complicated form.

(ii)Put (I — y³)^I= z, so that v3 = I — z³. Then_s(1 — y') -1 (— 3^y2) ^dy = dz;

i.e.—

(1 –

y3)

^-3

dy = dz.

—dz

—zdz

)

(I —

_z3)

which is not immediately integrable.

(iii)
Trials (i) and (ii) fail because the integrand does not contain a term y²dy which is the differential of 3y3. In dealing with functions of the form 1/a² + x²we have found that the substitution x = I/y changes

256 INTEGRAL CALCULUS

the index of the term outside the radical, and this suggests that the
above integral might be simplified by the substitution y = I/z, so that
dy = (— r/z²) dz.

Carrying out this substitution, we obtain
I    dz
I = —    —    dz    - - ( z² dz
= J
• ^z2(^I- 1/z'³)^l    z (^z3- I)*    I z³(^z3- 1)*
Since the integrand now consists of the two parts (i) z² dz, the dif

ferential of -iz³, and (ii) z³ (z³ — I)I, a function of *z³, the required conditions are satisfied. The substitution z³ = x will therefore simplify the integral.

A better substitution, which will rationalize the denominator of the integrand, is z³ — I = x³: this substitution will not affect the above con
ditions,
Putting z³ — I = x³, we have 3z² dz = 3x² dx, and
x²dx