How to integrate ∫(cos3x)^2?

using cos 2y = 2 cos^2y - 1

you get putting y = 3x

cos 6x = 2 cos ^2 3x - 1

or cos^2 3x= (cos 6x + 1)/2

now knowing integral of cos nx to be sin nx/n you get integrating

(1/6 sin 6x)/2 + 1/2 x

the constant of integration need to be added to get 1/12 sin 6x + x/2 + C