Trying to do the indefinate integral of x^3 * (9-x^2)^1/2 by using the substitution of x = 3 sin u I get 81* integral[ (sin^3 u)(cos u)] but again I appear to be wrong.
$\displaystyle x = 3\sin u $
$\displaystyle dx = 3\cos u \: du$
$\displaystyle \int {\color{red}x^{3}} {\color{blue}\sqrt{9 -x^{2}}} {\color{magenta}dx}$
$\displaystyle = \int {\color{red}\left(3\sin u\right)^{3}} {\color{blue}\sqrt{9 - 9\cos^{2} u}} \cdot {\color{magenta}3 \cos u \: du}$
$\displaystyle = \int 27 \sin^{3} u \cdot 3 \sqrt{1 - \cos^{2} u} \cdot 3\cos u \: du$
$\displaystyle = 243 \int \sin^{4} u \cdot \cos u \: du$
I think you forgot to make the sub for dx. This last integral should be straightforward. Make the substitution s = sin u and you should be on your way