Question

How do you find the integral of `int [cos^3 (2x)] dx`?

Answer 1

`1/2sin 2x-1/6sin^3 2x+C`

Explanation:

`I=int cos^3 2xdx = int cos 2x cos^2 2x dx`
`I=int cos 2x (1-sin^2 2x)dx`

`sin 2x=t => 2cos 2xdx = dt => cos 2xdx = dt/2`

`I=int(1-t^2)dt/2=1/2(t-t^3/3)+C`

`I=1/2sin 2x-1/6sin^3 2x+C`