5.3. INTEGRATION BY SUBSTITUTION
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easily evaluate the indefinite integral in u, and then go on to determine the desired overall
antiderivative of f ′ (g(x))g ′ (x). We call this process u-substitution. To see u-substitution at
work, we consider the following example.
Example 5.2. Evaluate the indefinite integral
x
3 · sin(7x
4 + 3) dx
and check the result by differentiating.
Solution. We can make two key algebraic observations regarding the integrand, x 3 ·
sin(7x 4 + 3). First, sin(7x 4 + 3) is a composite function; as such, we know we’ll need a
more sophisticated approach to antidifferentiating. Second, x 3 is almost the derivative
of (7x 4 + 3); the only issue is a missing constant. Thus, x 3 and (7x 4 + 3) are nearly
a function-derivative pair. Furthermore, we know the antiderivative of f (u) = sin(u).
The combination of these observations suggests that we can evaluate the given indefinite
integral by reversing the chain rule through u-substitution.
Letting u represent the inner function of the composite function sin(7x 4 + 3), we have
u = 7x 4 + 3, and thus
du
dx = 28x 3 . In differential notation, it follows that du = 28x 3 dx, and
thus x 3 dx =
1
28 du. We make this last observation because the original indefinite integral
may now be written
sin(7x
4 + 3) · x
3 dx,
and so by substituting the expressions in u for x (specifically u for 7x 4 + 3 and
1
28 du for
x 3 dx), it follows that
sin(7x
4 + 3) · x
3 dx =
sin(u) ·
1
28
du.
Now we may evaluate the original integral by first evaluating the easier integral in u,
followed by replacing u by the expression 7x 4 + 3. Doing so, we find
sin(7x
4 + 3) · x
3 dx =
sin(u) ·
1
28
du
=
1
28
sin(u) du
=
1
28
(− cos(u)) + C
= −
1
28
cos(7x
4 + 3) + C.