Is this an elementary integral? In a certain integration I stumbled upon an integral which I'm not sure is simple and elementary.
I wonder if it is one which is easily solvable or something which requires advanced tools to do: 
$$\int r^3\sqrt{8-r^2}dr$$ 
 A: Substitute $r=2\sqrt{2} \sin\theta$;
$dr=2\sqrt{2}cos\theta\  d\theta$
$$(2\sqrt{2})^5\int\sin^3\theta\ \cos^2 \theta \ d\theta=(2\sqrt{2})^5\int\sin^3\theta\ (1-\sin^2 \theta )\ d\theta$$$$(2\sqrt{2})^5\int(\sin^5 \theta- \sin^3 \theta )\ d\theta=(2\sqrt{2})^5\left(\int((1-\cos^2\theta)d(\cos\theta)-\int(1-\cos^2\theta)^2 d(\cos\theta)\right)$$
A: Let $u=8-r^2, du=-2rdr$ to get 
$\displaystyle\int r^3\sqrt{8-r^2}\,dr=-\frac{1}{2}\int r^2\sqrt{8-r^2}(-2r)dr=-\frac{1}{2}\int(8-u)\sqrt{u}du=-\frac{1}{2}\int(8u^{1/2}-u^{3/2})du$
$\displaystyle=-\frac{1}{2}\left[\frac{16}{3}u^{3/2}-\frac{2}{5}u^{5/2}\right]+C=-\frac{8}{3}(8-r^2)^{3/2}+\frac{1}{5}(8-r^2)^{5/2}+C$
A: Hint:
$$\int r^3\sqrt{8-r^2}dr= \frac12\int r^2\sqrt{8-r^2}d(r^2)=\frac12\int t\sqrt{8-t}\,dt.$$
You can integrate by parts on the second factor.

This substitution will work for all integrals of the form
$$\int P_o(r)(a^2-r^2)^\alpha dr$$ where $P_o$ is a polynomial with odd powers, and successive integrations by part will progressively lower the degree of the polynomial.
