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CHAPTER XVIII
INTEGRAL CALCULUS. METHODS OF INTEGRATION
I. THE METHOD OF SUBSTITUTION
98    INTEGRAL CALCULUS
 METHODS OF INTEGRATION    99
    ~)    sinxdx
f a+bcosx
Put    a+bcosx=y.
Then    — b sin xdx = dy.
Therefore
f    sin xdx __1 dy_ 1    log(a+bcosx)
J a+bcosx    bf y --blogy+c=—    b    +c.
    ~~)    dx _    dx
sin x        x
2 sin 5 cos 2
x sect dx
J 2tan
x
Put    tan 9 = y.
Then    i sec'- - dx = d y.
Therefore    f dx = f dye=logy+c
sin x


=logtan2+c.
Corollary. Since
cos x = sin    +xl,
J dx =1o, tan ('r + xl + c. cosx    a    4 2!
11. RATIONALISATION BY TRIGONOMETRICAL TRANSFORMATION
3. The above methods can be extended by the use of trigonofunctions and the relations which exist between them. The following are examples:
    f(1)    dx
'Va--x'
Let    x=asinv.
Then    dx = a cos ydy,
and    1/a — x2 = a cosy.
7—2

100    INTEGRAL CALCULUS
Therefore    (    _dx    = f dy = y + c
J N/a'—x'
x
=sin—1—+c. a
(2)         dx
f ~cz—a'
Let    x =asecy.
Then    dx = a sec y tan ydy,
and    'x2 — a2 = a tan y.
Therefore
J dx    = f    dy =1og tan ('r + y) + o
~/x = a2 J cosy    \4 2
1+tanY
= log —+o 1—tang
log 1 +siny+c
g cos y
1 +,1 a2 = log    
x+0
a x
=log(x+ N/x2—a2)
+c.
\    a
(3)    f      dx
N,/x2+a'.
Let    x = a tan y.
Then    dx = a sect ydy,
and    x2 + a2 = a sec y.
Therefore    / d
f 1/xd+ a2 — f cosy _ log tan (4 + 2) + c
=1og(sec y+tan y)+c [See Ex. (2)]
=log(x+ N/x2+a')+c. \    a

METHODS OF INTEGRATION    101
From the above integral, two other important forms can be obtained.
(4)    f    dx    
1 'Ja+2bx+cx2
This can be written as
    dx    _    d (cx + b)
. /(cx+b)'2+ac-bz    ,Vc{(cx+b)'+ac-b2}
V    c
Disregarding for the moment the constant multiplier, this can clearly be transformed into one of the forms given in Examples (1)—(3) above.
Corollary.
dx    =2log(x-a+,x-b)+c.
J '/(x - a) (x - b)
dx    2d Aix - a    /x - a
    Also         =2siri 1, b -a+c.
    V (x-a) (b-x)    (b - a) - (fix — a)z
f    (p + qx) dx    rq(b+cx)+(p~-qb)
    (5)    - c    c    dx
~a + 2bx + cx2    's/a + 2bx + cx2
- q f d (a + 2bx + cx2) + (pc - qb)    dx
2cJ .N/a+2bx+cx2    c    v'a+2bx+cx2
The value of the first integral is 2 (a + 2bx + cx2)l and the second integral can be dealt with as in Example (4).
4. General remarks in regard to the method of substitution. When the precise substitution which will enable the integral to be solved is not apparent the device illustrated in the following examples is often of assistance.
(1)    (x+a)dx
(x+b)vlx+c
The appropriate substitution is not apparent, so assume Nx+c=y'',
where r may have any value, to be determined subsequently.

102    INTEGRAL CALCULUS
Then    x+c=y2r
and    dx = 2ry2T-'dy.
The integral thus becomes
1(2r+a-c) try2r-'dy
(y2r + b - c) yr
Clearly this integral can be evaluated if yr = y2r-', i.e. if r = 1, when we get
f. (y2+a-c)di =2 ~[1 + (a-b)    d
J (y2+b-c)    (y+b-c) y
    -2y+ (a-b)    1
J    1
N/c-b _y-Vc-b - y+vc-b d
y 2y+ye-b logy+'c-b+k,
where y = 'x + c.
dx
(2)
(x p 'l(a + 2bx + cx2) '
Let us take the linear function as likely to lead to simpler results and, as before, put
x-p=yr
so that    dx = ry''-' dy.
Then the transformed integral is
r    ryr-'dy    f    rdy    
yr1/a+2b(yr+p)+c(yr+p)2    yva+2b(yr+p)+c(yr+p)2.
Obviously if r = - 1, the denominator reduces to the form
tiia'+2b'y+c'it
and the expression can be immediately integrated.
(3)    dx
•
(a + cx2)3
This example is a little more difficult than the two previous ones. As before, let x= yr, so that dx = ryr-' dy. Then we have
f dx _ ryr-' dy (a+cx2)l J(a+cy)1

METHODS OF INTEGRATION    103
The expression within the brackets in the denominator is of an even degree in y, so we will express the integral in the form
f ryr_2d(y2) _ f    ryr-2d(y2)    
2 (a + cy2')1    2ysr (ay 2r + c)
This can clearly be integrated if r — 2 = 3r, when r = — 1. The integral then becomes
— 1 I'    d.y2 1 f d (aye + c) 1
+k
2    (aye+c)- 2aJ (aye+c)1 a(ay2+c)y
   
x
+ k.
a (a + cx2)4

104    INTEGRAL CALCULUS
or    fudxdx=uv -Tvdudx.
Consequently an integral of a function of the form u dx can always be made to depend upon that of a function of the form v d.
The latter may prove to be a standard form and the use of the formula given above may enable an apparently intractable exto be integrated at once.
The advantage of the method is best exhibited by applying it to a few elementary cases.
 J\/a2 METHODS OF INTEGRATION    105
+x2 dx=f /a2+x2 xdx
=xa2+x2— fx"dxx2dx


I x2dx =x a2+x2— va2+x'
/_        Ir        rye
=xN a2+x2—I+x2—/     ~dx
,a2 + x2
f a2        f a2dx
,\/a2 + x2


Therefore 2 f A/a2 +x2dx = x ~a2 + x2 + a' log [x + a'2 + x] + c, and    f/a2 + x2 dx =2 /a2+x'+ 22 log [x+/a2+x2]+ .
(5) flog(x + 'X2 + a2)dx = f log(x+ x2+a') dx
xlog(x+x2+(1,2)— ('xdlog(x+d    +a')dx
=xlog(x+~x2+a2)— r xdx 'Vx2+a'
=xlog(x+Vx2+a2)—Vx'+a2+c.


7. The above process can be expressed in a general form as follows:

futvtdt =futd(dtdI)dt
=ut fv,dt— f ut(fvtdt)dt.
fb
To apply this formula to the definite integral I utvtdt, we can
a
t
write fvdt as (vkdk, the upper limit being taken as t since the
a
integral must itself be a function of t.

106    INTEGRAL CALCULUS
Thus we get
    pb    b    t    ~dzt    , t
bu=J utvtdt = ut~ vkdk — dtt (. a vkdk) dt]
    a    a_    a f vkdk —ua~a vkdk— fbdut(j vkdk) dt
a    a    a dt    a
= ub ~bvkdk — ~b dut (~tvkdk) dt,
a    a d t    a
—f a since vkdk = 0. Alternatively, if we express fvt cit as the definite integral J vkdk, we obtain
t
butvtdt= ua bvkdk +f bdttt Cl bvkdkI
.a    .a    a dt    t
 METHODS OF INTEGRATION    107
Thus the degree of the term (1 — x)' has been reduced by unity. Successive applications of the formula will reduce the degree further, until, if n is an integer, the term ultimately equals (1 — x)° or unity.
In the special case where the value of this integral is taken between the limits 0 and 1, we have
    1x'"(1—x)ndx= [m     (1 — x)' + .0m+1 xn(1—x)n-'dx
o    m n
-' (1 —    dx,
ni + 1
whence by successive applications of the formula we find ultiif n be an integer that
f 1 x'n (1 — x). dx =    n ! m ! [1 xm+n dx =    n ! m,    
o    n+m! 0    n+m+1!'
Many other important integrals can be dealt with in a similar way.
IV. INTEGRATION BY THE USE OF PARTIAL FRACTIONS
10. Any expression of the form F~xx)
, where F(x) and f x) are
both rational integral algebraic functions of x, can be expressed as the sum of a number of terms of which the general forms are arx'' and    For if the degree of F (x) is equal to or greater than
(xC,.~r.
that of f (x), by division we can obtain a quotient, together with a new fraction in which the numerator is of lower degree than the denominator. The quotient provides the terms of the first form and the new fraction can be split up by partial fractions into terms of the second form. The integrals of these general forms are known and consequently the whole expression can be integrated.
The following are examples:
f (1)    (p+qx)dx    1    aq+p—bq+p dx
(x—a) (x—b) a—bJ [x — a x —b
1
=a   b[(aq+p)log(x—a)—(bq+ p) log (x—b)]+o.
    clx    _ 1    (Ix
(2)    J a~~ + 2bx + c a    b 2 ac — L'
x+a) +    a2 —

108    INTEGRAL CALCULUS
The form of the integral depends upon whether ac – b2 is positive or negative.
In the former case the integral clearly is equal to 1–      tan—'
ax+b
ac b2    v'ac – b2
In the latter case the integral becomes 1    to ax + b -1lb2 – ac
2,Jb2–ac gax+b+A.lb2–ac
fo(x+l)(x+2) x'dx    = ~i    1    8
(3)    o x–3–x+1+x+ 2 dx
= l 2 -3x—1og(x+1)+8log(x+2)~1 o
=8log3–9log2–.