Calculus- Early Transcendentals, 2021a

260 Techniques of Integration
and
∫
−3cos2xdx= − 3 2 sin2x
are easy. The cos 3 2x integral is like the previous example:
∫
∫
−cos 3 2xdx = −cos2xcos 2 2xdx
∫
= −cos2x(1 − sin 2 2x)dx
∫
= − 1 2 (1 − u2 )du
) (u − u3
= − 1 2 3
= − 1 (
)
sin2x − sin3 2x
.
2
3
And finally we use another trigonometric identity, cos 2 x =(1 + cos(2x))/2:
∫
∫ 1 + cos4x
3cos 2 2xdx= 3
dx = 3 (
x + sin4x )
.
2 2 4
So at long last we get
∫
sin 6 xdx= x 8 − 3
16 sin2x − 1 (
)
sin2x − sin3 2x
+ 3 (
x + sin4x )
+C.
16
3 16 4
Next, we turn our attention to products of secant and tangent. Some we already know how to do.
∫
∫
sec 2 xdx= tanx +C
secxtanxdx= secx +C
♣
We can also integrate tanx quite easily.
Example 7.16: Integrating Tangent
∫
Evaluate tanxdx.
Solution. Note that tanx = sinx and let u = cosx, sothatdu = −sinxdx.
cosx
∫
∫ sinx
tanxdx =
cosx dx Rewriting tanx
∫ sinx du
=
Using the substitution
u −sinx
1
= −∫
du Cancelling and pulling the −1 out
u ∫ 1
= −ln|u| +C Using formula dx = ln|u| +C
u
= −ln|cosx| +C Replacing u back in terms of x
= ln|secx| +C Using log properties and secx = 1/cosx