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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:

$$\int r^3\sqrt{8-r^2}dr$$

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  • $\begingroup$ Is $r=x? Also, skip the _ if you do not have limits. $\endgroup$ – mickep Jul 1 '16 at 9:00
  • $\begingroup$ fuc%%% up is not really in the SE spirit. Anyhow, the integral is trivial once you substitute $r=\sqrt{8}\sin{x}$ $\endgroup$ – b00n heT Jul 1 '16 at 9:13
  • $\begingroup$ This is not an answer but is related to your question. You may find this article interesting. It is Brian Conrad's "Impossibility theorems for elementary integration". $\endgroup$ – MPW Jul 1 '16 at 18:44
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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.

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Substitute $r=2\sqrt{2} \sin\theta$; $dr=2\sqrt{2}cos\theta\ d\theta$ $$(2\sqrt{2})^5\int\sin^3\theta\ \cos^2 \theta \ d\theta=(2\sqrt{2})^5\int\sin^3\theta\ (1-\sin^2 \theta )\ d\theta$$$$(2\sqrt{2})^5\int(\sin^5 \theta- \sin^3 \theta )\ d\theta=(2\sqrt{2})^5\left(\int((1-\cos^2\theta)d(\cos\theta)-\int(1-\cos^2\theta)^2 d(\cos\theta)\right)$$

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Let $u=8-r^2, du=-2rdr$ to get

$\displaystyle\int r^3\sqrt{8-r^2}\,dr=-\frac{1}{2}\int r^2\sqrt{8-r^2}(-2r)dr=-\frac{1}{2}\int(8-u)\sqrt{u}du=-\frac{1}{2}\int(8u^{1/2}-u^{3/2})du$

$\displaystyle=-\frac{1}{2}\left[\frac{16}{3}u^{3/2}-\frac{2}{5}u^{5/2}\right]+C=-\frac{8}{3}(8-r^2)^{3/2}+\frac{1}{5}(8-r^2)^{5/2}+C$

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