Question

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

Answer 1

`intsin^2xcos^2xdx=x/8-(sin4x)/32`

Explanation:

As `sin2x=2sinxcosx`, `sinxcosx=1/2sin2x` and

`sin^2xcos^2x=1/4sin^2 2x=1/4(1/2(1-cos4x))=(1-cos4x)/8`

Hence `intsin^2xcos^2xdx=int(1-cos4x)/8 dx`

= `x/8-1/8intcos4xdx`

Let `u=4x`, then `du=4dx`

and `intcos4xdx=1/4intcosudu=1/4sinu=1/4sin4x`

Hence `intsin^2xcos^2xdx=x/8-1/8xx1/4sin4x=x/8-(sin4x)/32`