Question

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

Answer 1

`F(x) = -x/2-1/8sin4x+C`

Explanation:

First we note that:

`cos^2(2x) = (cos(4x)-1)/2`

So, its primitive can be computed as:

`int_0^x cos^2(2t)dt = 1/2int_0^x (cos(4t)-1)dt = -x/2 + 1/8int_0^x cos(4t)d(4t) = -x/2-1/8sin4x+C`