I need to integrate:

$\displaystyle

\int \cot^{5}\left(3x\right)dx

$

I started to set u=3x then 1/3du=dx and I get

$\displaystyle

\frac{1}{3}\int \cot^{5}\left(u\right)dx

$

I know that

$\displaystyle

\frac{d}{dx} \cot x = -\csc^{2} x

$

and that

$\displaystyle

\cot^{2} x = \csc^{2} x -1

$

But now I am not sure how to go on. In the case of cot^3 I would write it as cotx and cot^2 x and set t=cotx.

Since I have cot^5 I do not know what to do.

Thanks for any help.