Question

How do you find the antiderivative of `(cos(3x))^3`?

Answer 1

`1/3sin3x-1/9sin^3 3x+c`

Explanation:

`intcos^3 3xdx`

`=intcos3xcos^2 3xdx`

`=intcos3x(1-sin^2 3x)dx`

`I=int(cos3x-cos3xsin^2 3x)dx`

now
`d/(dx)(sin^3 3x)=3 xx3xxsin^2 3xcos3x=9sin^2 3xcos3x`

`I=1/3sin3x-1/9sin^3 3x+c`