Calculus- Early Transcendentals, 2021a

7.1. Substitution Rule 253
This is a problem because our integrals can’t have a mixture of two variables in them. Usually this means
we chose our u incorrectly. However, in this case we can eliminate the remaining x’s from our integral by
using:
u = x 2 + 1 → x 2 = u − 1.
We get:
∫
∫
x 2 u 1/2 du = (u − 1)u 1/2 du
=
∫
u 3/2 − u 1/2 du
= 2 5 u5/2 − 2 3 u3/2 +C
= 2 5 (x2 + 1) 5/2 − 2 3 (x2 + 1) 3/2 +C
♣
The next example shows how to use the Substitution Rule when dealing with definite integrals.
Example 7.8: Substitution Rule
Evaluate
∫ 4
2
xsin(x 2 )dx.
Solution. First we compute the antiderivative, then evaluate the integral. Let u = x 2 ,sodu = 2xdx or
xdx= du/2. Then
∫
xsin(x 2 )dx =
∫ 1
2 sinudu= 1 2 (−cosu)+C = −1 2 cos(x2 )+C.
Now
∫ 4
2
xsin(x 2 )dx = − 1 2 cos(x2 )
∣
4
2
= − 1 2 cos(16)+1 2 cos(4).
A somewhat neater alternative to this method is to change the original limits to match the variable u. Since
u = x 2 ,whenx = 2, u = 4, and when x = 4, u = 16. So we can do this:
∫ 4
∫ 16
∣
xsin(x 2 1
∣∣∣
16
)dx =
2
4 2 sinudu= −1 2 (cosu) = − 1
4
2 cos(16)+1 2 cos(4).
An incorrect, and dangerous, alternative is something like this:
∫ 4
2
xsin(x 2 )dx =
∫ 4
2
∫ 4
∣
1
∣∣∣
4
2 sinudu= −1 2 cos(u)
2
= − 1 2 cos(x2 )
∣
4
2
= − 1 2 cos(16)+1 2 cos(4).
1
This is incorrect because sinudu means that u takes on values between 2 and 4, which is wrong. It
2 2
is dangerous, because it is very easy to get to the point − 1 ∣ ∣∣∣
4
2 cos(u) and forget to substitute x 2 back in for
2