How prove this $P_{n}(x)=\frac{1}{\pi}\int_{0}^{\pi}\frac{dt}{(x-\sqrt{x^2-1}\cos{t})^n}dt$ let $x>1$ and $n\in N$,show that
the  function 
$$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}(x+\sqrt{x^2-1}\cos{t})^ndt$$
is polynomial and the dgree is $n$.and we have
$$P_{n}(x)=\dfrac{1}{\pi}\int_{0}^{\pi}\dfrac{1}{(x-\sqrt{x^2-1}\cos{t})^n}dt$$
maybe this is Legendre Polynomial,But I can't.
 A: First Integral
Substitute $t\mapsto\frac\pi2-t$, then exploit the oddness of $\sin(t)$:
$$
\begin{align}
\int_0^\pi(x+\sqrt{x^2-1}\cos(t))^n\,\mathrm{d}t
&=\int_{-\pi/2}^{\pi/2}(x+\sqrt{x^2-1}\sin(t))^n\,\mathrm{d}t\\
&=\int_{-\pi/2}^{\pi/2}\sum_{k=0}^n\binom{n}{k}x^{n-k}\left(\sqrt{x^2-1}\sin(t)\right)^k\,\mathrm{d}t\\
&=\int_{-\pi/2}^{\pi/2}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{n-2k}\left(\sqrt{x^2-1}\sin(t)\right)^{2k}\,\mathrm{d}t\\
&=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{n-2k}(x^2-1)^k\frac\pi{4^k}\binom{2k}{k}
\end{align}
$$
and each term of the last sum is a polynomial of degree $n$ with positive lead coefficient.

Second Integral
$$
\begin{align}
\pi I_n(x)
&=\int_0^\pi\frac{\mathrm{d}t}{(x-\sqrt{x^2-1}\cos(t))^{n+1}}\\
&=\frac1{x^{n+1}}\int_{-\pi/2}^{\pi/2}\frac{\mathrm{d}t}{(1-\sqrt{1-1/x^2}\sin(t))^{n+1}}\\
&=\frac1{x^{n+1}}\int_{-\pi/2}^{\pi/2}\sum_{k=0}^\infty\binom{-n-1}{k}\left(-\sqrt{1-1/x^2}\sin(t)\right)^k\,\mathrm{d}t\\
&=\frac1{x^{n+1}}\int_{-\pi/2}^{\pi/2}\sum_{k=0}^\infty\binom{-n-1}{2k}\left((1-1/x^2)\sin^2(t)\right)^k\,\mathrm{d}t\\
&=\frac1{x^{n+1}}\sum_{k=0}^\infty\binom{-n-1}{2k}(1-1/x^2)^k\frac\pi{4^k}\binom{2k}{k}\\
&=\frac\pi{x^{n+1}}\sum_{k=0}^\infty\binom{n+2k}{2k}\sqrt{\frac{1-1/x^2}{4}}^{\,\large2k}\binom{2k}{k}\\
\end{align}
$$
Setting $u=\sqrt{\frac{1-1/x^2}{4}}$, we have $\displaystyle\sum_{n=0}^\infty I_n(x)v^n$ is
$$
\begin{align}
\frac1x\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{n+2k}{2k}\binom{2k}{k}u^{2k}(v/x)^n
&=\frac1x\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{n+2k}{n}\binom{2k}{k}u^{2k}(v/x)^n\\
&=\frac1x\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{-2k-1}{n}\binom{2k}{k}u^{2k}(-v/x)^n\\
&=\frac1{x-v}\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{u}{1-v/x}\right)^{2k}\\
&=\frac1{x-v}\left(1-4\left(\frac{u}{1-v/x}\right)^2\right)^{-1/2}\\
&=\frac1{\sqrt{1-2vx+v^2}}\\
&=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{v(2x-v)}{4}\right)^k
\end{align}
$$
The terms with $v^n$ appear when $k\le n\le2k$. Thus, $x^k$ appears in $I_n(x)$ when $n/2\le k\le n$. Therefore, $I_n(x)$ is a degree $n$ polynomial and the coefficient of $x^n$ in $I_n(x)$ is $\binom{2n}{n}2^{-n}$.
A: For the first integral, the result is 
$$\left(x-\sqrt{x^2-1}\right)^n \, _2F_1\left(\frac{1}{2},-n;1;2-2 x
   \left(x+\sqrt{x^2-1}\right)\right)$$  
For the second integral, the result is  
$$\left(\sqrt{x^2-1}+x\right)^{-n} \, _2F_1\left(\frac{1}{2},n;1;2 x
   \left(\sqrt{x^2-1}-x\right)+2\right)$$  
As Norbert said, they are not the same. For the values Norbert used, the values are respectively $17$ and $11/2$  
May be, there are some typo's somewhere. Please check.
A: To show it's polynomial:
Use binomial theorem to $(x+\sqrt{x^2-1}\cos{t})^n$.
Reduction formula for $A_n = \int_{0}^{\pi}(\cos{t})^ndt$ that could be useful:
$A_n = \frac{n-1}{n}A_{n-2}$
