Integrating $\int\frac{x^3}{\sqrt{9-x^2}}dx$ via trig substitution What I have done so far:
Substituting $$x=3\sin(t)\Rightarrow dx=3\cos(t)dt$$ converting our integral to 
$$I=\int\frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{27\sin^3(t) dt}{3\sqrt{\cos^2(t)}}3\cos(t)dt\\
\Rightarrow \frac{I}{27}=\int \sin^3(t)dt=-\frac{\sin^2(t)\cos(t)}{3}+\frac{2}{3}\int\sin(t)dt\\
\Rightarrow I=18\cos(t)-9\sin^2(t)\cos(t)+c$$
Which is where I'm stuck. I can substitute back in for the $\sin(t)$ terms but don't know hot to deal with the $\cos(t)$ terms and back out information about $x$. 
edited: because I miswrote the problem, I am sorry.

Substitution $x=3\sin{y} \rightarrow dx=3\cos{y}$ $dy$
Then 
$27 \int \frac{\sin^3{y}}{3\cos{y}}.3\cos{y}$ $dy$
$=\frac{27}{4}\int 4\sin^3{y}$ $dy$
Use Identity 
$4\sin^3{y}=3\sin{y}-\sin{3y}$
$\frac{27}{4}\int 4\sin^3{y}$ $dy$ $=\frac{27}{4}\int 3\sin{y}-\sin{3y}$ $dy$
$=\frac{27}{4}(-3\cos{y}-\frac{\cos{3y}}{3})$
From substitution,
$\cos{y}=\frac{3}{\sqrt{9-x^2}}$
$\cos{3y}=4\cos^3{y}-3\cos{y}$
$\therefore \frac{27}{4}(-3\cos{y}-\frac{\cos{3y}}{3})=\frac{27}{4}[-\frac{9}{\sqrt{9-x^2}}-\frac{1}{3}(\frac{108}{(9-x^2)^{\frac{3}{2}}}-\frac{9}{\sqrt{9-x^2}})]$
You should be able to carry forward.Good luck!
