Prove $ \int\limits_{0}^{+\infty}{\frac{\mathrm{d}x}{\cosh^{n}{x}}}=\int\limits_{0}^{\frac{\pi}{2}}{\cos^{n-1}{x}\,\mathrm{d}x} $ Let $ n\geq 1 $, Denoting 
$$ F_{n}=\int\limits_{0}^{+\infty}{\frac{\mathrm{d}x}{\cosh^{n}{x}}},\>\>\>\>\>\>W_{n}=\int\limits_{0}^{\frac{\pi}{2}}{\cos^{n}{x}\,\mathrm{d}x} $$
How would you prove that $ \left(\forall n\geq 1\right),\ F_{n}=W_{n-1} $, $\textbf{without}$ finding closed forms for any of $ F_{n} $ and $ W_{n} \cdot $

I know Futuna's integrals and Wallis's integrals are great classics, and finding closed forms for each of them would be child's play, but I haven't yet found a way to prove the nice relationship between them without looking for their closed forms.

 A: Note that we can use the Beta Function as a means to proceed.

Accordingly, we first use the well-known representations of the Beta Function, $$B(x,y)=2\int_0^{\pi/2}\sin^{2x-1}(\theta)\cos^{2y-1}(\theta)\,d\theta$$ and 
$$B(x,y)=\int_0^\infty\frac{t^{x-1}}{(1+t)^{x+y}}\,dx$$ to write
$$\begin{align}
\int_0^{\pi/2} \cos^{n-1}(x)\,dx&=\frac12B\left(\frac12,\frac n2\right)\\\\
&=\frac12\int_0^\infty \frac{t^{-1/2}}{(1+t)^{(n+1)/2}}\,dt
\end{align}$$

Now, let $t=\sinh^2(x)$ so that $dt=2\sinh(x)\cosh(x)\,dx$.  Then, we have
$$\begin{align}
\frac12\int_0^\infty \frac{t^{-1/2}}{(1+t)^{(n+1)/2}}\,dt&=\frac12\int_0^\infty\frac{\frac1{\sinh(x)}}{\cosh^{n+1}(x)}\,2\sinh(x)\cosh(x)\,dx\\\\
&=\int_0^\infty \frac{1}{\cosh^n(x)}\,dx
\end{align}$$
And we are done!
A: There is in fact a "simple" substitution that suffices for this: the Gudermannian function. In particular, this satisfies
$$ \DeclareMathOperator{gd}{gd} \DeclareMathOperator{sech}{sech} \gd 0 = 0 , \quad \gd x \to \pi/2 \text{ as } x \to \infty , $$
and most importantly,
$$ \gd' x = \sech x , \quad \cos(\gd x ) = \sech x , $$
whence we put $t = \gd x$ in $W_{n-1}$, to find
$$ W_{n-1} = \int_0^{\pi/2} \cos^{n-1} \theta \, d\theta = \int_0^{\infty} (\cos(\gd x))^{n-1} \sech x \, dx = \int_0^{\infty} \sech^n x \, dx = F_n , $$
as required.
A: Substitute $  \cos x = \text{sech} \ t$ to have 
$$dx = \frac { \text{sech} t\tanh t}{\sin x}dt = \frac{dt}{\cosh t}$$
Thus,
$$\int_0^{\frac{\pi}{2}}{\cos^{n-1}{x}dx}
= \int_0^{\infty} \frac{dt}{\cosh^n t}$$
