Question

How do you find the antiderivative of `cos^(2)3x dx`?

Answer 1

The answer is `=(sin6x+6x)/12+C`

Explanation:

We use `cos2x=2cos^2x-1`
Adapting this
`cos6x=2cos^2(3x)-1`
Therefore, `Cos^2(3x)=(cos6x+1)/2`
So `intCos^2(3x)dx=int((cos6x+1)dx)/2`
`=1/2(sin(6x)/6+x)+C`
`(sin6x+6x)/12+C`