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.

Since
$$
\frac{\sin(x)}{\cosh(x)+\cos(x)}=2\sum_{n\ge 1}(-1)^{n-1}e^{-nx}\sin(nx
)$$
(shown here), we can combine the above with $\sin^2(x)\sin(nx) = \frac14 \left[2 \sin(n x) + \sin((2  - n) x) - \sin((2 + n) x)\right]$ and $\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}\mathrm{d}x = \arctan\left(\frac{b}{a}\right)$ to get
\begin{align*}
I &= \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1}\int_{0}^{\infty}\frac{e^{-nx}}{x}\left[2 \sin(n x) + \sin((2  - n) x) - \sin((2 + n) x)\right] \, \mathrm{d}x\\
& = \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1} \left[\frac\pi2 + \arctan\left(\frac{2-n}{n} \right) - \arctan\left(\frac{2+n}{n} \right)\right]\\
& = \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1}  \left[\arctan(n+1) - \arctan(n-1)\right]\\
& = \frac{1}{2}\left[-\arctan(0) + \arctan(1) \right]\\
 & = \frac{\pi}{8}
\end{align*}
as desired.

Note that to re-write the arctangent expressions from equations $2\to 3$ we combine $\arctan(\alpha) - \arctan(\beta) = \arctan\left(\frac{\alpha - \beta}{1 + \alpha\beta} \right)$ and $\arctan(x) + \arctan\left( \frac1x\right)= \frac\pi2,\ x>0$. Thus
\begin{align*}
\frac\pi2 + \arctan\left(\frac{2-n}{n} \right) - \arctan\left(\frac{2+n}{n} \right) &=\frac\pi2 + \arctan\left(\frac{-n^2}{2}\right) \\
&\overset{\color{blue}{n^2/2 >0}}{=}  \arctan\left(\frac{2}{n^2}\right) \\
&=\arctan(n+1) - \arctan(n-1)
\end{align*}
also exploiting the fact that $\arctan(x)$ is odd.
