Using $\displaystyle x=3sin\theta$

$\displaystyle \int x^3 \sqrt {9-x^2} ~dx$

I think I solved the part under the radical correctly ($\displaystyle 3cos\theta$)

If I plug in $\displaystyle x=3sin\theta$ and $\displaystyle dx=3cos\theta$ and use $\displaystyle 3cos\theta$ for the radical I get:

$\displaystyle \int 243sin^3\theta cos^2\theta ~d\theta$

I can change it to $\displaystyle 243\int sin\theta(1-cos\theta)(cos^2\theta) $

But don't know what to do next.