Complicated trigonometric integral proof $$r>1$$
The following is applying gauss' law by explicit integration.
$$\int_0^{\pi } \frac{\sin (\theta )}{\sqrt{r^2-2 r \cos (\theta )+1}} \, d\theta=\frac{2}{r}$$
The following is finding the potential of a charged spherical shell by explicit integration. Mathematica takes forever to solve this and I cannot figure out how it would be done.
$$\int _0^{2 \pi }\int _0^{\pi }\frac{\sin (\theta ) \cos (\theta )}{\sqrt{r^2-2 r (\cos (\psi )\cos (\theta ) +\sin (\psi )\sin (\theta )  \cos (\phi ))+1}}d\theta d\phi=\frac{4 \pi }{3}\frac{\cos(\psi)}{r^2}$$
 A: We begin with the integral $I(r,\psi)$ given by
$$I(r,\psi)=\int_{0}^{2\pi}\int_{0}^{\pi}\frac{\sin \theta \cos \theta}{\sqrt{r^2-2r\left(\cos \theta \cos \psi+\sin \theta \sin \psi \cos \phi\right)+1}}\,d\theta\,d\phi$$
Using spherical harmonics, we can expand the denominator of the integrand as 
$$\frac{1}{\sqrt{r^2-2r\left(\cos \theta \cos \psi+\sin \theta \sin \psi \cos \phi\right)+1}}=4\pi\sum_{\ell=0}^{\infty}\sum_{m=-\ell}^{\ell}\frac{1}{2\ell +1}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}Y_{\ell m}(\theta,\phi)Y_{\ell m}(\psi,0)$$
The integration over $\phi$ is trivial and reveals
$$\begin{align}
I(r,\psi)&=2\pi \int_{0}^{\pi}\sin \theta \cos \theta \sum_{\ell=0}^{\infty}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}P_{\ell}(\cos \theta)P_{\ell}(\cos \psi)\,d\theta\\\\&=2\pi \sum_{\ell=0}^{\infty}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}P_{\ell}(\cos \psi)\int_{0}^{\pi}\sin \theta \cos \theta P_{\ell}(\cos \theta)\,d\theta \tag 1
\end{align}$$
Noting that $P_1(\cos \theta)=\cos \theta$ and exploiting the orthogonality of the Legendre Polynomials, we obtain
$$\begin{align}
I(r,\psi)&=2\pi\sum_{\ell=0}^{\infty}\frac{r_{<}^{\ell}}{r_{>}^{\ell+1}}P_{\ell}(\cos \psi)\left(\frac{2}{2\ell +1}\delta_{\ell,1}\right)\\\\
&=\frac{4\pi}{3}\frac{r_{<}}{r_{>}^{2}}P_{1}(\cos \psi)\\\\
&=\frac{4\pi}{3}\frac{r_{<}}{r_{>}^{2}}\cos \psi\\\\
\end{align}$$
For $r>1$, $r_{<}=1$ and $r_{>}=r$.  Finally, we have
$$\bbox[5px,border:2px solid #C0A000]{I(r,\psi)=\frac{4\pi}{3} \frac{\cos \psi}{r^2}}$$
A: Substitution Method
Let $r^2-2r\cos \theta+1=u^2\implies 2r\sin\theta d\theta=2udu$ or $\sin\theta d\theta=\frac{udu}{r}$
$$\int_{0}^{\pi}\frac{\sin\theta}{\sqrt{r^2-2r\cos \theta+1}}d\theta=\int_{r-1}^{r+1}\frac{1}{u}\frac{udu}{r}$$
$$=\frac{1}{r}\int_{r-1}^{r+1}du=\frac{2}{r}$$
