what is the integral of cos^2 3x dx ?

We have to integrate (cos 3x)^2
dx.

We know that cos 2x = 2*(cos x)^2 -
1

=> (cos x)^2 = (1 + cos
2x)/2

The expressions we have can be written as (1 + cos
6x)/2

Int [ (cos 3x)^2
dx)

=> Int [ (1 + cos 6x)/2
dx]

let u = 6x

=> du/6
= dx

Int [ (1 + cos 6x)/2
dx]

=> (1/2)(1/6)*Int [ (1 + cos u) du
]

=> (1/12)* (u + sin
u)

substituting u =
6x

=> (1/12)(6x + sin 6x) +
C

The required integral is (1/12)(6x + sin
6x) + C