Computation of $\int_0^{\pi} \frac{\sin^n \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} \, d\theta$ 
Show that 
  $$\begin{align*} \forall x \in [-1,1]: \int_0^{\pi} \frac{\sin^n \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} \, d\theta &= c_n \tag{1} \\ \int_0^{\pi} \frac{\sin^{n+2} \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} \, d\theta &= a_n \cdot x^2+b_n \tag{2} \end{align*}$$ where $a_n, b_n, c_n$ are constants (which do not depend on $x \in [-1,1]$), $n \geq 3$.

I tried several approaches (differentiation to show that the derivative of $(1)$ is equal to $0$, Weierstraß substitution, ...), but always got stuck. For example, one can show that the integrand in $(1)$ equals 
$$\begin{align*} \frac{\sin^n \theta}{(1+x^2-2x \cdot \cos \theta)^{\frac{n}{2}}} &= \left( \sqrt{ \left( \frac{x-\cos \theta}{\sin \theta} \right)^2+1} \right)^{-n} \\ &= \left( \sqrt{4 \left( \frac{\sin \frac{\theta+\varrho}{2} \cdot \sin \frac{\theta-\varrho}{2}}{\sin \theta}  \right)^2+1} \right)^{-n} \end{align*}$$ where $x=\cos \varrho$. I hoped to get some kind of symmetrization out of it, but (as far as I can see) it doesn't work.
Any ideas? (The aim is to find a rather quick or direct proof - a lengthy one is already known, using a recursive/inductive approach.)
Thanks!
 A: The Gegenbauer polynomials were built to solve problems like this.
Suppose $m\in\mathbb{N}$.
We have
$$\begin{eqnarray*}
\int_0^{\pi} d\theta \,
    \frac{\sin^{n+2m} \theta}{(1-2x \cos \theta+x^2)^{n/2}}
&=& \int_0^\pi  d\theta \, \sin^{n+2m} \theta
    \sum_{k=0}^\infty C_k^{(n/2)}(\cos\theta) x^k \\
&=& \int_{-1}^1  d u \, (1-u^2)^{n/2-1/2} (1-u^2)^m
    \sum_{k=0}^\infty C_k^{(n/2)}(u) x^k
    \hspace{5ex} (u=\cos\theta) \\
&=& \int_{-1}^1  d u \, (1-u^2)^{n/2-1/2}
    \sum_{l=0}^m \beta_{2l} C_{2l}^{(n/2)}(u)
    \sum_{k=0}^\infty C_k^{(n/2)}(u) x^k \\
&=& \sum_{l=0}^m \sum_{k=0}^\infty  \beta_{2l} x^k
    \underbrace{\int_{-1}^1  d u \, (1-u^2)^{n/2-1/2}
    C_{2l}^{(n/2)}(u) C_k^{(n/2)}(u)}_{\gamma_{2l}^{(n/2)} \delta_{2l,k}} \\
&=& \sum_{l=0}^m \beta_{2l} \gamma_{2l}^{(n/2)} x^{2l}.
\end{eqnarray*}$$
This proves the claim for any $m\in\mathbb{N}$. 
The restriction $n\ge 3$ is too loose.
The result holds for any $n$ such that $n>-1$.
For a given $m$ it is a straightforward exercise to find the $\beta$s and $\gamma$s.
For example, using the fact that $1=C_0^{(n/2)}(u)$ and the normalization relation for the Gegenbauer polynomials we find
$$\int_0^{\pi} d\theta \, \frac{\sin^{n} \theta}{(1-2x \cos \theta+x^2)^{n/2}}
= \frac{\sqrt{\pi}\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}.$$
Some details:


*

*$1/(1-2x \cos \theta+x^2)^{n/2}$
is the generating function for the Gegenbauer polynomials.

*The polynomials have the parity property.

*The polynomials provide an orthogonal basis for functions on $[-1,1]$ with weight function $(1-u^2)^{n/2-1/2}$.

A: You are in the good way.
The integral can be expressed as
$$
I(\varphi)=\int_{0}^{\pi}f(\theta,\varphi)d\theta
$$
 where
$$ f(\theta,\varphi)=\left[1+\left(\dfrac{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)}{\sin\theta}\right)^{2}\right]^{-\frac{n}{2}}
$$
 where $x=\cos\varphi$. 
Now, $f(\theta,\varphi)=f(-\theta,\varphi)$, so is no harm in writing the integral as
$$
I(\varphi)=\dfrac{1}{2}\int_{-\pi}^{\pi}f(\theta,\varphi)d\theta
$$
We want to prove that
$$
 \cfrac[l]{\partial I(\varphi)}{\partial\varphi}=\dfrac{1}{2}\int_{-\pi}^{\pi}\cfrac[l]{\partial f(\theta,\varphi)}{\partial\varphi}d\theta=0
$$
so it is enough to prove that
$$
 g(\theta,\varphi)=\cfrac[l]{\partial f(\theta,\varphi)}{\partial\varphi}
$$
is antisymmetrical in $\theta$.
Deriving we obtaing
$$
g(\theta,\varphi)=\dfrac{n\left(\dfrac{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)}{\sin\theta}\right)}{\left[1+\left(\dfrac{2\sin\left(\dfrac{\theta+\varphi}{2}\right)\sin\left(\dfrac{\theta-\varphi}{2}\right)}{\sin\theta}\right)^{2}\right]^{\frac{n}{2}+1}}
$$
and we see that  $g(\theta,\varphi)=-g(-\theta,\varphi)$.
For expression (2), the proof is analogous.
