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How do you find the integral of #sin^2 (x)cos^2 (x) dx#?

1 Answer

Jul 19, 2016

#1/32(4x-sin4x)+C#.

Explanation:

Let us note that, #sin^2xcos^2x=1/4(sin2x)^2=sin^2(2x)/4#

Recall that, #sin^2theta=(1-cos2theta)/2#

Letting #theta=2x#, we have, #sin^2(2x)=(1-cos4x)/2#

#:. intsin^2xcos^2xdx=1/8int(1-cos4x)dx=1/8(x-sin(4x)/4)=1/32(4x-sin4x)+C#.

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