We can write ∫cos³3xdx as:

∫cos3x(cos²3x)dx and

further as:

=cos3x(1-sin²3x)dx

=∫cos3xdx-∫cos3x(sin²3x)dx

Taking A=∫cos3xdx

Solving for A

A=

Taking B=∫cos3x(sin²3x)dx

In this taking sin3x=t

Differentiating on both sides we get

3cos3xdx=dt

Solving by putting these values we get

B

Substituting values we get

B

Our final answer is A+B i.e