# What is indefinite integral of cos^2(3x)?

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We have to find the indefinite integral of (cos 3x)^2

cos 2x = 2*(cos x)^2 - 1

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

Int[(cos 3x)^2 dx]

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

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

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

let y = 6x, dy/6 = dx

=> x/2 + Int[(cos y)/12 dy]

=> x/2 + sin y/12

=> x/2 + (sin 6x)/12 + C

**The required integral is x/2 + (sin 6x)/12 + C**

To calculate the indefinite integral of the given function, we'll have to recall the half angle identity:

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

We'll evaluate the indefinite integral:

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

We'll apply the property of integral to be additive:

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

Int (1 + cos 6x)dx/2 = x/2 + sin 6x/12 + C

**The indefinite integral of the given function is: F(x)=x/2 + sin 6x/12 + C.**