Calculus- Early Transcendentals, 2021a

7.1. Substitution Rule 251
Example 7.4: Substitution
∫
Evaluate the following integral:
2x
√
1 − 4x 2 dx.
Solution. We try the substitution:
u = 1 − 4x 2 .
Then,
du = −8x dx
In the numerator we have 2x dx, so rewriting the differential gives:
− 1 du = 2xdx.
4
Then the integral is:
∫
2x
√
1 − 4x 2 dx = ∫ (1
− 4x
2 ) −1/2
(2xdx)
=
∫
u −1/2 (
− 1 4 du )
=
( ) −1 u
1/2
4 1/2 +C
√
1 − 4x 2
= − +C
2
♣
Example 7.5: Substitution
∫
Evaluate the following integral:
cosx(sinx) 5 dx.
Solution. In this question we will let u = sinx. Then,
du = cosx dx.
Thus, the integral becomes:
∫
cosx(sinx) 5 dx =
∫
u 5 du
= u6
6 +C
= (sinx)6
6
+C