Showing $\pi\int_{0}^{\infty}[1+\cosh(x\pi)]^{-n}dx={(2n-2)!!\over (2n-1)!!}\cdot{2\over 2^n}$ Showing 

$$\pi\int_{0}^{\infty}[1+\cosh(x\pi)]^{-n}dx={(2n-2)!!\over (2n-1)!!}\cdot{2\over 2^n}\tag1$$

Recall
$$1+\cosh(x\pi)={(e^{x\pi}+1)^2\over 2e^{x\pi}}\tag2$$
$$I_n=2^n\pi\int_{0}^{\infty}{e^{xn\pi}\over (1+e^{x\pi})^{2n}}dx\tag3$$
$$I_n={2^n\pi\over n\pi}\int_{0}^{\infty}n\pi e^{nx\pi}(1+e^{x\pi})^{-2n}dx\tag4$$
$$I_n=\left.{2^n\over n}\cdot{1\over (1+e^{x\pi})^{2n-1}}\right|_{0}^{\infty}\tag5$$
$$I_n={1\over n(2n-1)2^{n-1}}\tag6$$
Help, where did I went wrong?

New Edit
From (3) we make a substitution
let $u=x\pi \rightarrow du=dx$
$$I_n=2^n\int_{0}^{\infty}e^{un}(1+e^u)^{-2n}du\tag{3a}$$
 A: First: remove the useless constant by setting $x=\frac{z}{\pi}$. Then, through $z=\log u$ and $v=\frac{1}{u}$:
$$ \int_{0}^{+\infty}(1+\cosh(z))^{-n}\,dz = 2^n\int_{1}^{+\infty}\frac{\left(2+u+\frac{1}{u}\right)^{-n}}{u}\,du=2^n\int_{0}^{1}\frac{\left(2+v+\frac{1}{v}\right)^{-n}}{v}\,dv$$
so the LHS equals:
$$ 2^{n-1}\int_{0}^{+\infty}\frac{u^{n-1} du}{(u+1)^{2n}}\,du = 2^n\int_{0}^{+\infty}\frac{t^{2n-1}\,dt}{(1+t^2)^{2n}}= 2^{n-1} B(n,n) = 2^{n-1}\frac{\Gamma(n)^2}{\Gamma(2n)}.$$
As an alternative, just apply IBP multiple times. It leads to a recursion similar to the one for $\int_{0}^{\pi/2}\sin^{2n}(\theta)\,d\theta.$
A: Take $u=\tanh\left(x\right)
 $, then $\cosh\left(x\right)=\frac{1+u^{2}}{1-u^{2}}
 $ and $dx=\frac{2du}{1-u^{2}}
 $ so $$\int_{0}^{\infty}\frac{1}{\left(\cosh\left(x\right)+1\right)^{n}}dx=\int_{0}^{1}\frac{2}{\left(1-u^{2}\right)\left(\frac{1+u^{2}}{1-u^{2}}+1\right)^{n}}du
 $$ $$=\frac{1}{2^{n-1}}\int_{0}^{1}\left(1-u^{2}\right)^{n-1}du=\frac{1}{2^{n}}\int_{0}^{1}\left(1-v\right)^{n-1}v^{-1/2}dv=\frac{B\left(1/2,n\right)}{2^{n}}$$ $$=\frac{1}{2^{n}}\frac{\sqrt{\pi}\left(n-1\right)!}{\Gamma\left(n+1/2\right)}.$$ 
A: Let $I(n)$ be the integral given by 
$$\begin{align}
I(n)&=\pi\int_0^\infty\frac{1}{\left(1+\cosh(\pi x)\right)^n}\,dx\\\\
&=\int_0^\infty\frac{1}{\left(1+\cosh(x)\right)^n}\,dx\tag 1
\end{align}$$
Then, using the identity $1+\cosh(x)=\frac12 e^{x}\left(1+e^{-x}\right)^2$ in $(1)$ reveals
$$\begin{align}
I(n)&=2^n\int_0^\infty \frac{e^{-nx}}{\left(1+e^{-x}\right)^{2n}}\,dx\\\\
&=2^n\int_0^1 \frac{x^{n-1}}{(1+x)^{2n}}\,dx\tag 2
\end{align}$$
Now, making the substitution $x\to 1/x$ in $(2)$, we obtain
$$\begin{align}
I(n)&=2^n\int_1^\infty \frac{x^{n-1}}{(1+x)^{2n}}\,dx \tag 3
\end{align}$$
Combining $(2)$ and $(3)$ yields
$$I(n)=2^{n-1}\int_0^\infty \frac{x^{n-1}}{(1+x)^{2n}}\,dx \tag 4$$
Then, enforcing the substitution $x\to \frac{x}{1-x}$ in $(4)$, we find that
$$\begin{align}
I(n)&=2^{n-1}\int_0^1 x^{n-1}(1-x)^{n-1}\,dx\\\\
&=2^{n-1}B(n,n)\\\\
&=2^{n-1}\frac{\Gamma^2(n)}{\Gamma(2n)}\\\\
&=\frac{(n-1)!}{(2n-1)!!}\\\\
&=\frac{(2n-2)!!}{2^{n-1}(2n-1)!!}
\end{align}$$
as was to be shown!
