Question

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

Answer 1

`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`.