Reference request or advices for aficionados to find information about $\int_0^\pi\frac{(\cos(x))^{2n}}{e^{\cos(x)}-1}dx$, and a related question While I was playing with Wolfram Alpha online calculator I wondered about a possible closed-form of the first few cases for integers $n\geq 1$ in 
$$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx.\tag{1}$$
My problem (as aficionado) here that I think that it could be in the literature, and I am curious (today and sometimes in past ocassions about series and integrals) to know how to find some of my creations.

Question 1. Do you know how to find if the integral $$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx$$ is in the literature? I am asking about the tools for an aficionado, where I should to search, with what key words in Internet? What sites? Optionally if you know this integral feel free to add the reference answering this question as a reference request, and I search and read such closed-form from the literature. Many thanks.

I'm almost sure that I've seen this kind of integrals, or that should be easy to get, but I don't remember where. I know (without a justification) cases as $n=1,2$ and $n=3$.

Question 2. Calculate a good approximation of $$\int_0^\pi\frac{\cos^{3}(x)}{e^{\cos(x)}-1}dx.\tag{2}$$
  Provide your justification. Many thanks.

There were a typo, see the first comment.
 A: Q1. This integral has closed form and for integer $n$, it reduces to a simple Beta function $\left(B\right)$
$$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx=\int_0^{\pi/2}\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx+\int_{\pi/2}^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx \tag{1}$$
$$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx=\int_0^{\pi/2}\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx+\int_{0}^{\pi/2}\frac{(-\cos(x))^{2n}}{e^{-\cos(x)}-1}dx \tag{2}$$
$$\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}+\frac{(-\cos(x))^{2n}}{e^{-\cos(x)}-1}=-\cos(x)^{2n} \quad n\in \mathcal{Z} \tag{3}$$
$$\int_0^\pi\frac{\cos^{2n}(x)}{e^{\cos(x)}-1}dx=-\int_0^{\pi/2}\cos^{2n}(x)dx=-\frac{1}{2}B\!\left(\frac{1}{2},n+\frac{1}{2}\right)=-\frac{\pi\left(2n\right)!}{n!^{2}\,2^{2n+1}} \tag{4}$$
Q2. More generally, term wise integration of the generating function of the Bernoulli polynomials, with $b_{n}$ the $n^{th}$ Bernoulli number, gives:
$$ \int_{0}^{\pi}\!{\frac { \cos^v \left( x \right)}{{
{\rm e}^{\cos \left( x \right) }}-1}}\,{\rm d}x=-\frac{\left( 1+ \left( -1 \right) ^{v} \right)}{4}\,B\! \left( \frac{1}{2}
,\frac{v+1}{2} \right)   +\frac{\left( 1- \left( -1 \right) ^{v} \right)}{2}\,
  \sum _{n=0}^{\infty }{\frac 
{b_{2n} \,B\! \left( \frac{1}{2},n+\frac{v}{2} \right) }{
 \left( 2\,n \right) !}}
 $$
$$\tag{5}$$
hence:
$$\int_{0}^{\pi}\!{\frac { \cos^3 \left( x \right) }{{
{\rm e}^{\cos \left( x \right) }}-1}}\,{\rm d}x=\pi\sum _{n=0}^{\infty}{\frac 
{ \left( 2\,n+1 \right)  \left( n+1 \right) b_{2n} }{{2}^{2\,n+1} \left(  \left( n+1 \right) ! \right) ^{2}}
}\approx {\frac {58705}{110592}}\pi \quad\quad\text{(4 terms)} \tag{6}
$$
