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]
...
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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