<s>[INST] Seeking for help to find a formula for $\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}$, where $a>1.$ When tackling the question, I found that for any $a>1$,
$$
I_1(a)=\int_{0}^{\pi} \frac{d x}{a-\cos x}=\frac{\pi}{\sqrt{a^{2}-1}}.
$$
Then I started to think whether there is a formula for the integral
$$
I_n(a)=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}, $$
where $n\in N.$
After trying  some substitution and integration by parts, I still failed and got no idea for reducing the power n. After two days, the Leibniz Rule for high derivatives come to my mind.
Differentiating $I_1(a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\displaystyle \begin{array}{l}
\displaystyle \int_{0}^{\pi} \frac{(-1)^{n-1}(n-1) !}{(a-\cos x)^{n}} d x=\frac{d^{n-1}}{d a^{n-1}}\left(\frac{\pi}{\sqrt{a^{2}-1}}\right) \\ \displaystyle 
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{(-1)^{n-1} \pi}{(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{1}{\sqrt{a^{2}-1}}\right) \tag{*}\label{star}
\end{array}
$$
I am glad to see that the integration problem turn to be merely a differentiation problem.
Now I am going to find the $(n-1)^{th} $ derivative by Leibniz Rule.
First of all, differentiating $I_1(a)$ w.r.t. $a$ yields $$
\left(a^{2}-1\right) \frac{d y}{d a}+a y=0 \tag{1}\label{diffeq}
$$
Differentiating \eqref{diffeq} w.r.t. $a$ by $(n-1)$ times gets $$
\begin{array}{l}
\displaystyle \left(a^{2}-1\right) \frac{d^{n} y}{d a^{n}}+\left(\begin{array}{c}
n-1 \\
1
\end{array}\right)(2 a) \frac{d^{n-1} y}{d a^{n-1}}+2\left(\begin{array}{c}
n-1 \\
2
\end{array}\right) \frac{d^{n-2} y}{d a^{n-2}}+x \frac{d^{n-1} y}{d a^{n-1}}+(n-1) \frac{d^{n-2} y}{d a^{n-2}}=0
\end{array}
$$
Simplifying, $$
\left(a^{2}-1\right) y^{(n)}+(2 n-1) ay^{(n-1)}+(n-1)^{2} y^{(n-2)}=0 \tag{2}\label{diffrec}
$$
Initially, we have $ \displaystyle y^{(0)}=\frac{1}{\sqrt{a^{2}-1}}$ and $ \displaystyle y^{(1)}=-\frac{a}{\left(a^{2}-1\right)^{\frac{3}{2}}}.$
By \eqref{diffrec}, we get $$
y^{(2)}=\frac{2 a^{2}+1}{\left(a^{2}-1\right)^{\frac{5}{2}}}
$$ and $$
\displaystyle  y^{(3)}=-\frac{3 a\left(2 a^{2}+3\right)}{\left(a^{2}-1\right)^{\frac{7}{2}}}
$$
Plugging into \eqref{star} yields $$
\begin{aligned}
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{3}} &=\frac{\pi}{2} y^{(2)}=\frac{\pi\left(2 a^{2}+1\right)}{2\left(a^{2}-1\right)^{\frac{5}{2}}} \\
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{4}} &=-\frac{\pi}{6} \cdot \frac{3 a\left(2 a^{2}+3\right)}{\left(a^{2}-1\right)^{\frac{7}{2}}} =-\frac{\pi a\left(2 a^{2}+3\right)}{2\left(a^{2}-1\right)^{\frac{7}{2}}}
\end{aligned}
$$
Theoretically, we can proceed to find $I_n(a)$ for any $n\in N$ by the recurrence relation in $(2)$ .
By Mathematical Induction, we can further prove that the formula is $$
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{\pi P(a)}{\left(a^{2}-1\right)^{\frac{2 n-1}{2}}}
$$
for some polynomial $P(a)$ of degree $n-1$.
Last but not least, how to find the formula for $P(a)$?  Would you help me?
 [/INST] Express the integral as
\begin{align}
\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}
=\sum_{k=0}^{[\frac{n-1}2]} \frac{\binom{n-1}{2k}a^{n-2k-1}}{(a^2-1)^{n-1/2}}\int_0^{\pi}\cos^{2k}x \ dx
= \frac{\pi P(a)}{\left(a^{2}-1\right)^{n-{1}/{2}}}
\end{align}
which leads to
$$P(a) = \sum_{k=0}^{[\frac{n-1}2]}
\binom{n-1}{2k} \frac{(2k-1)!!}{(2k)!!} a^{n-2k-1}
$$
</s>