Integrating $r^4\cos^2\theta\sin\theta$ Over an Ellipse I was tutoring someone and one of the problems was to integrate $r^4\cos^2\theta\sin\theta$ over the portion of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ that lies in the first quadrant.  So, $0\leq \theta\leq\frac{\pi}{2}$.  But, given a $\theta$, $r$ runs from $0$ to
$$
\sqrt{\frac{36+36\tan^2\theta}{9+4\tan^2\theta}}.
$$
After integrating with respect to $r$ first, it looks very nasty.
Is there some clever method, such as a change of variables, that makes this easy to integrate?  

The integral yields to standard substitutions. Sometime when I am feeling in a clever mood I may come up with a quick to verify solution. At this point, I am interested in showing that the integral can be done by absolutely routine substitutions. By analyzing the substitutions, we could come up with shortcuts 
We will not pay attention to multiplicative constants, since we would probably get them wrong anyway. After the integration with repsect to $r$, we end up with an $r^5$ term. But $r^2=36/(9\cos^2\theta+4\sin^2\theta)$. So we end up wanting something that (apart from multiplicative constants) looks like 
$$\int_0^{\pi/2}\frac{\cos^2\theta\sin\theta}{(9\cos^2\theta+4\sin^2\theta)\sqrt{9\cos^2\theta +4\sin^2\theta}}d\theta.$$
Replace $9\cos^2\theta+4\sin^2\theta$ by $4+5\cos^2\theta$ and make the substitution $u=\cos\theta$. We end up wanting an integral of the shape
$$\int_0^1 \frac{u^2}{(4+5u^2)^2\sqrt{4+5u^2}}du.$$
In this situation, it is standard to make a substitution of the shape $\sqrt{5}u=2\tan\phi$, or  something similar with hyperbolic sine. After a little work we end up with something that, apart from constants, looks like the absolutely familiar 
$$\int \sin^2\phi\cos\phi\; d\phi.$$
