<s>[INST] 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.
 [/INST] A nice trick to do in this situation is to write 
$$
\sin^3t = (1-\cos^2t)\sin t.
$$
You can now do a secondary $u$-substitution where you can take $u = \cos t$ and $du = -\sin tdt$.
Edit
Where you left off you had
$$
\frac{I}{27} \;\; =\;\; \int \sin^3t dt \;\; =\;\; \int (1-\cos^2t)\sin t dt
$$
If we now take the $u$-substitution I suggested above we find
$$
\frac{I}{27} \;\; =\;\;- \int (1 - u^2) du \;\; =\;\; \frac{1}{3}u^3 - u + C
$$
for some constant $C$.  We can do the first re-substitution
$$
\frac{I}{27} \;\; =\;\; \frac{1}{3} \cos^3t - \cos t+C
$$
and now you can do the other substitution by noting that $\cos t = \sqrt{1 - \sin^2t} = \sqrt{1 - \frac{x^2}{9}} = \frac{1}{3} \sqrt{9 - x^2}$.
Additional Edit
If we plug back in the above expressions we get
\begin{eqnarray*}
\frac{I}{27} & = & \cos t\left (\frac{1}{3}\cos^2t - 1\right ) + C \\
& = & \frac{1}{3} \sqrt{9-x^2}\left (\frac{1}{27}(9-x^2) - 1\right ) + C\\
I & = & \frac{1}{3} \sqrt{9-x^2} (9- x^2 - 27) + C \\
& = & -\frac{1}{3} \sqrt{9-x^2}(x^2 + 18) + C.
\end{eqnarray*}
The answer you saw on Wolfram Alpha was the result of taking through the calculation as far as it could go. 
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