How to show $\int_0^{\infty} \frac{\sin^3 x}{\cosh x\>+\>\cos x}\frac{dx}x=\frac{\pi}{8}$ I would like to evaluate the integral below$$\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x $$
which I found to be $\frac \pi8$ numerically. I was able to evaluate a similarly looking, yet simpler, integral
$$\int_0^{\infty} \frac{\sin x}{\cosh x+\cos x}\frac{dx}x =\frac\pi4$$
by writing the integrand in the Frullani format and apply the theorem. However, it does not work in this case and I am not sure if they are related. Any solution is appreciated.

It was shown in this answer that if $n$ is a nonnegative integer, then $$\int_{0}^{\infty} \frac{\sin\left((2n+1)x \right)}{\cosh (x) + \cos (x)} \, \, \frac{\mathrm dx}{x} = \frac{\pi}{4}. $$
Therefore, $$\begin{align} \int_{0}^{\infty}\frac{\sin^{3}(x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} &= \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin (x)- \sin(3x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} \\ & = \frac{1}{4} \left(\frac{3 \pi}{4}- \frac{\pi}{4} \right) \\ & = \frac{\pi}{8}. \end{align}$$
In general, if $m$ is a nonnegative integer, then $$ \begin{align} \int_{0}^{\infty}\frac{\sin^{2m+1}(x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} &\overset{(1)}{=} \frac{1}{2^{2m}} \sum_{k=0}^{m} (-1)^{m-k} \binom{2m+1}{k} \int_{0}^{\infty}\frac{\sin \left((2m+1-2k)x \right)}{\cosh(x) + \cos (x)}\frac{\mathrm dx}{x} \\ &= \frac{\pi}{2^{2m+2}}\sum_{k=0}^{m} (-1)^{m-k} \binom{2m+1}{k}. \end{align}$$
$(1)$ https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae
