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5.3. INTEGRATION BY SUBSTITUTION
To check our work, we observe by the Chain Rule that
d
dx
−
1
28
cos(7x
4 + 3) + C
= −
1
28
· (−1) sin(7x
4 + 3) · 28x
3 = sin(7x
4 + 3) · x
3
,
which is indeed the original integrand.
An essential observation about our work in Example 5.2 is that the u-substitution only
worked because the function multiplying sin(7x 4 + 3) was x 3 . If instead that function
was x 2 or x 4 , the substitution process may not (and likely would not) have worked. This
is one of the primary challenges of antidifferentiation: slight changes in the integrand
make tremendous differences. For instance, we can use u-substitution with u = x 2 and
du = 2xdx to find that
xe
x 2 dx =
e
u ·
1
2
du
=
1
2
e
u du
=
1
2
e
u + C
=
1
2
e
x 2
+ C.
If, however, we consider the similar indefinite integral
e
x 2 dx,
the missing x to multiply e x 2 makes the u-substitution u = x 2 no longer possible. Hence,
part of the lesson of u-substitution is just how specialized the process is: it only applies to
situations where, up to a missing constant, the integrand that is present is the result of
applying the Chain Rule to a different, related function.
Activity 5.8.
Evaluate each of the following indefinite integrals by using these steps:
• Find two functions within the integrand that form (up to a possible missing
constant) a function-derivative pair;
• Make a substitution and convert the integral to one involving u and du;
• Evaluate the new integral in u;
• Convert the resulting function of u back to a function of x by using your earlier
substitution;