Calculus- Early Transcendentals, 2021a

258 Techniques of Integration
= u7
7 − 2u9
9 + u11
11 +C
= sin7 x
7
− 2sin9 x
9
+ sin11 x
11 +C
Solution 2: Since the power of cosine is odd, we use the substitution u = sinx and du = cosxdx,that
is, dx =
du
cosx .
∫
Then sin 6 xcos 5 xdx is equal to:
∫
=
∫
=
∫
=
∫
=
∫
=
u 6 cos 5 x
du
cosx
u 6 ( cos 2 x ) 2
du
u 6 (1 − sin 2 x) 2 du
u 6 (1 − u 2 ) 2 du
u 6 − 2u 8 + u 10 du
Using the substitution
Canceling a cosx and rewriting cos 4 x
Using trig identity cos 2 x = 1 − sin 2 x
Writing integral in terms of u’s
Expand and collect like terms
= u7
7 − 2u9
9 + u11
+C Integrating
11
= sin7 x
7
− 2sin9 x
9
+ sin11 x
11 +C Replacing u back in terms of x ♣
Example 7.13: Odd Power of Cosine
∫
Evaluate cos 3 xdx.
Solution. Since the power of cosine is odd, we use the substitution u = sinx and du = cosxdx.Thismay
seem strange at first since we don’t have sinx in the question, but it does work!
∫
cos 3 xdx =
=
=
=
∫
∫
∫
∫
cos 3 du
x
cosx
cos 2 xdu
Using the substitution
Canceling a cosx
(1 − sin 2 x)du Using trig identity cos 2 x = 1 − sin 2 x
(1 − u 2 )du Writing integral in terms of u’s
= u − u3
+C Integrating
3
= sinx − sin3 x
+C Replacing u back in terms of x
3