Question

What is the integral of `int (cos2x)^2 dx`?

Answer 1

`= 1/8 sin 4x + x/2 + C`

Explanation:

use the ID: `cos 2A = 2 cos^2 A - 1` or `cos^2 A = 1/2(cos 2A + 1)`

the integral becomes

`1/2 int cos4x + 1 \ dx`

`= 1/2 (1/4 sin 4x + x) + C`

`= 1/8 sin 4x + x/2 + C`