Proving that an integral is zero in order to prove Newton's shell theorem Prove that for any $a\in(-1,1)$
$$\int_{0}^{\pi}\frac{(\cos t-a)\sin t}{\left(1+a^2-2a\cos t\right)^{3/2}}\,dt = 0.$$

The first stage would be to replace: $$x = \cos t - a, dx = \sin t dt$$
This leads to:
$$\int_{-1 -a}^{1 -a}\frac{-x\,dx}{\left(1-a^2-2ax\right)^{3/2}} = 
\left. \frac{(a^2+ax-1)dx}{a^2\left (1-a^2-2ax\right)^{1/2}} \right|_{-1-a}^{1-a}=0$$
