Evaluate∫ sin4x cos2xdx
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∫sin
4
x⋅cos
2
xdx=∫(sin
2
x)(sin
2
x⋅cos
2
x)dx
=∫
2
1
(1−cos2x)(
2
sin2x
)
2
dx
cos2x=1−2sin
2
x
sin
2
x=
2
1−cos2x
sin2x=2sinxcosx
The integral becomes
=∫
2
1
(1−cos2x)⋅(
2
sin2x
)
2
dx
=
8
1
(∫(1−cos2x)⋅sin(2x)dx))
=
8
1
(∫(sin
2
2x−cos2x⋅sin
2
(2x))dx)
=
8
1
∫
2
1
⋅2sin
2
(2x)dx−∫
2
1
⋅2cos2xsin
2
2xdx
=
8
1
(∫
2
1
⋅(1−cos(4x))dx−∫
2
1
sin
2
2xdsin2x)
=
16
1
(∫(1−cos(4x)dx−∫sin
2
(2x)dsin2x)
dx
dsin2x
=2cos2x
d
sin(2x)=2cos2x⋅dx
=
16
1
(x−
4
sin4x
−
3
sin
3
(2x)
)+C.
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