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Integrate $\int_{0}^{2\pi } \int_{0}^{a} \int_{0}^{\pi/2} \frac{ G\rho r^2 sin \theta}{ {(r^2-2rt cos \theta + t^2 )}^{\frac{1}{2}}} d \theta dr d \phi$, where $\rho, t $ are constants.

Sorry for the bad format. My idea was to integrate with respect to $r$ first, hence something of the form

$\int_{0}^{a} \frac{Ar^2}{\sqrt{(r-B)^2 + C}} dr $

But even in this simpler version I find it difficult as substitutions such as $r = B+ \sqrt{C}\tan \theta $ that get rids of the bottom still yields ugly trigonometric equations.

Any ideas? Thank you!

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For example, the inner integral is:

$$\int_0^{\pi/2}\frac{Gpr^2\sin\theta}{(r^2-2rt\cos\theta+t^2)^{1/2}}d\theta=\left.\frac{Gpr}{t}(r^2-2rt\cos\theta+t^2)^{1/2}\right|_0^{\pi/2}=$$$${}$$

$$=\frac{Gpr}t\left[(r^2+t^2)^{1/2}-(r^2-2rt+t^2)^{1/2}\right]=\frac{Gpr}t\left(\sqrt{r^2+t^2}-(r-t)\right)$$

You're left then with the middle integral, which is pretty simple since for example

$$\int r\sqrt{r^2+t^2}\;dr=\frac13(r^2+t^2)^{3/2}$$

and etc.

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