# Prove equality with product of improper integrals

I have to prove the following equality: $$\int\limits_0^{+\infty} \frac{dx}{\sqrt{\cosh x}} \cdot \int\limits_0^{+\infty} \frac{dx}{\sqrt{\cosh^3 x}} = \pi.$$

The first integral Wolfram Mathematica somehow evaluates to $$4\sqrt{\frac{2}{\pi}}\Gamma\left(\frac 54\right)^2.$$

But I don't know how to simplify it to such form and deal with the second integral.

We have: $$I_n = \int_{0}^{+\infty}\frac{dx}{\cosh^{n/2} x} = \int_{1}^{+\infty}\frac{dt}{t^{n/2}\sqrt{t^2-1}} =\int_{0}^{1}\frac{t^{n/2-1}}{\sqrt{1-t^2}}\,dt\tag{1}$$ so, by Euler's Beta function:
$$I_n = \frac{1}{2}\int_{0}^{1}t^{n/4-1}(1-t)^{-1/2}\,dt = \frac{\Gamma(n/4)\,\Gamma(1/2)}{2\, \Gamma(n/4+1/2)}=\frac{\sqrt{\pi}}{2}\cdot\frac{\Gamma(n/4)}{\Gamma(n/4+1/2)}\tag{2}$$ and through $\Gamma(z+1)=z\,\Gamma(z)$ we have:
$$I_1\cdot I_3 = \frac{\pi}{4}\cdot\frac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}=\color{red}{\pi}.\tag{3}$$