An identity between integral

Question

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

Henry Lee · Answer

$$frac{sin^ntheta}{(rho^2-2rhocostheta+1)^{n/2}}=frac{sin^nalpha}{rho^n}$$
also:
$$frac{costheta}{(rho^2-2rhocostheta+1)^{1/2}}-frac{rhosin^2theta}{(rho^2-2rhocostheta+1)^{3/2}}=frac{cosalpha}{rho}frac{dalpha}{dtheta}$$
so if we try and put some of this substitution in (not a complete integral yet) we have:
$$int_0^pifrac{sin^nalpha}{rho^n}frac{1}{sin^2theta}dtheta$$
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

An identity between integral

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