An identity between integral

Question

I want prove that, for $n\in\mathbb{N}$ and $\rho > 1$: $$ \int_{0}^{\pi}\!\!\! \frac{\sin^{n - 2\,}\left(\theta\right)} {\left[\rho^{2} - 2\rho\cos\left(\theta\right) + 1\right]^{n/2}\,}\,\mathrm{d}\theta = \frac{1}{\rho^{n - 2}\left(\rho^2-1\right)} \int_{0}^{\pi}\!\!\!\sin^{n - 2}\left(\theta\right) \,\mathrm{d}\theta $$ The hint is to use the following change of variable: $$ \frac{\sin\left(\theta\right)} {\left[\rho^{2} - 2\rho\cos\left(\theta\right) + 1\right]^{1/2}} = \frac{\sin\left(\alpha\right)}{\rho} $$ But i can't go on. Any help is appreciated.

Answer 1

$$\frac{\sin^n\theta}{(\rho^2-2\rho\cos\theta+1)^{n/2}}=\frac{\sin^n\alpha}{\rho^n}$$ also: $$\frac{\cos\theta}{(\rho^2-2\rho\cos\theta+1)^{1/2}}-\frac{\rho\sin^2\theta}{(\rho^2-2\rho\cos\theta+1)^{3/2}}=\frac{\cos\alpha}{\rho}\frac{d\alpha}{d\theta}$$ so if we try and put some of this substitution in (not a complete integral yet) we have: $$\int_0^\pi\frac{\sin^n\alpha}{\rho^n}\frac{1}{\sin^2\theta}d\theta$$ if you take a look at the derivative above it has a relationship to the original function, so using this should be very helpful. Good luck