How to prove $\int_{0}^{\pi}\frac{\sin^n{x}}{(1+r^2-2r\cos{x})^{(n+2)/2}}\,\mathrm dx=\frac{1}{1-r^2}\int_{0}^{\pi}\sin^n{x}\,\mathrm dx$
Prove that
$$\displaystyle\int_{0}^{\pi}\dfrac{\sin^n{x}}{(1+r^2-2r\cos{x})^{(n+2)/2}}\,\mathrm dx=\dfrac{1}{1-r^2}\int_{0}^{\pi}\sin^n{x}\,\mathrm dx,$$ where $n\ge 1, n\in \Bbb N,|r|<1.$
I think use $$\dfrac{1-r^2}{1-2r\cos{x}+r^2}=1+2\displaystyle\sum_{m=1}^{\infty}r^m\cos{(mx)},$$ where $|r|<1.$
But it is very ugly.
The integral in the LHS is a partial case of formula 3.665.2 from Gradshteyn & Ryzhik. The integral in the RHS also is well-known (ibid 3.621.1 ). I don't find any motivation to reinvent the wheel.