Question

What is the integral of `int sin^2 (x).cos^2 (x) dx`?

Answer 1

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

Explanation:

As `sin2x=2sinxcosx`

`intsin^2xcos^2xdx=1/4int(4sin^2xcos^2x)dx`

= `1/4intsin^2(2x)dx`

= `1/4int(1-cos4x)/2dx`

= `x/8-1/8intcos4xdx`

= `x/8-1/8xx(sin4x)/4+c`

= `x/8-(sin4x)/32+c`