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.
Best Answer
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!