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Over the next few sections we examine some techniques that are frequently successful when
seeking antiderivatives of functions. Sometimes this is a simple problem, since it will be
apparent that the function you wish to integrate is a derivative in some straightforward
way. For example, faced with Z
x10 dx
we realize immediately that the derivative of x11 will supply an x10 : (x11 )′ = 11x10 . We
don’t want the “11”, but constants are easy to alter, because differentiation “ignores” them
in certain circumstances, so
d 1 11 1
x = 11x10 = x10 .
dx 11 11
From our knowledge of derivatives, we can immediately write down a number of an-
tiderivatives. Here is a list of those most often used:
xn+1
Z
xn dx = + C, if n 6= −1
n+1
Z
x−1 dx = ln |x| + C
Z
ex dx = ex + C
Z
sin x dx = − cos x + C
163
,164 Chapter 8 Techniques of Integration
Z
cos x dx = sin x + C
Z
sec2 x dx = tan x + C
Z
sec x tan x dx = sec x + C
1
Z
dx = arctan x + C
1 + x2
1
Z
√ dx = arcsin x + C
1 − x2
8.1 Substitution
Needless to say, most problems we encounter will not be so simple. Here’s a slightly more
complicated example: find Z
2x cos(x2 ) dx.
This is not a “simple” derivative, but a little thought reveals that it must have come from
an application of the chain rule. Multiplied on the “outside” is 2x, which is the derivative
of the “inside” function x2 . Checking:
d d
sin(x2 ) = cos(x2 ) x2 = 2x cos(x2 ),
dx dx
so Z
2x cos(x2 ) dx = sin(x2 ) + C.
Even when the chain rule has “produced” a certain derivative, it is not always easy to
see. Consider this problem: Z p
x3 1 − x2 dx.
p
There are two factors in this expression, x3 and 1 − x2 , but it is not apparent that the
chain rule is involved. Some clever rearrangement reveals that it is:
1
Z p Z p
3
x 1− x2 dx = (−2x) − (1 − (1 − x2 )) 1 − x2 dx.
2
This looks messy, but we do now have something that looks like the result of the chain
√
rule: the function 1 − x2 has been substituted into −(1/2)(1 − x) x, and the derivative
, 8.1 Substitution 165
of 1 − x2 , −2x, multiplied on the outside. If we can find a function F (x) whose derivative
√
is −(1/2)(1 − x) x we’ll be done, since then
d 2 2 1 p
′
F (1 − x ) = −2xF (1 − x ) = (−2x) − (1 − (1 − x2 )) 1 − x2
dx 2
p
= x3 1 − x2
But this isn’t hard:
1 √ 1
Z Z
− (1 − x) x dx = − (x1/2 − x3/2 ) dx (8.1.1)
2 2
1 2 3/2 2 5/2
=− x − x +C
2 3 5
1 1
= x− x3/2 + C.
5 3
So finally we have
1 1
Z p
3 2
2
x 1 − x dx = (1 − x ) − (1 − x2 )3/2 + C.
5 3
So we succeeded, but it required a clever first step, rewriting the original function so
that it looked like the result of using the chain rule. Fortunately, there is a technique that
makes such problems simpler, without requiring cleverness to rewrite a function in just the
right way. It sometimes does not work, or may require more than one attempt, but the
idea is simple: guess at the most likely candidate for the “inside function”, then do some
algebra to see what this requires the rest of the function to look like.
One frequently good guess is any complicated expression inside a square root, so we
start by trying u = 1 − x2 , using a new variable, u, for convenience in the manipulations
that follow. Now we know that the chain rule will multiply by the derivative of this inner
function:
du
= −2x,
dx
so we need to rewrite the original function to include this:
√ −2x x2 √ du
Z p Z Z
3 3
x 1 − x2 = x u dx = u dx.
−2x −2 dx
Recall that one benefit of the Leibniz notation is that it often turns out that what looks
like ordinary arithmetic gives the correct answer, even if something more complicated is
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