TITLE: The entry-level PhD integral: $\int_0^\infty\frac{\sin 3x\sin 4x\sin5x\cos6x}{x\sin^2 x\cosh x}\ dx$
QUESTION [38 upvotes]: I hope you find this integral interesting.

Evaluate
  $$\int_0^\infty\frac{\sin\left(\,3x\,\right)\sin\left(\,4x\,\right)
\sin\left(\,5x\,\right)\cos\left(\,6x\,\right)}{x\,\sin^{2}\left(\,x\,\right)\cosh\left(\,x\,\right)}\,\,\mathrm{d}x\tag1$$

This problem is taken from the PhD graduate entry tests in my college. I've tried to use product-to-sum trigonometric identities
$$2\sin 4x\sin 3x=\cos x-\cos 5x$$
and
$$2\cos 6x\sin 5x=\sin 11x-\sin x$$
I got a bunch of the following form 
$$\int_0^\infty\frac{\sin \alpha x\cos \beta x}{x\sin^2 x\cosh x}\ dx\quad\Longrightarrow\quad\int_0^\infty\frac{\sin \gamma x}{x\sin^2 x\cosh x}\ dx\tag2$$
I tried
$$I'(\gamma)=\int_0^\infty\frac{\cos \gamma x}{\sin^2 x\cosh x}\ dx\tag3$$
but the latter form is not easy to evaluate either. Can anyone here help me to evaluate $(1)$? Thanks in advance.

REPLY [12 votes]: In fact, we have
$$
\begin{align}
I(M,N)&=\int_0^\infty\frac{\sin Nx\sin(N+1)x\sin Mx\cos(M+1)x}{x\sin^2 x\cosh x}\ dx\\[10pt]
&=\sum_{m=1}^M\sum_{n=1}^N\left[\arctan\left( e^{(m+n)\pi} \right)-\arctan\left( e^{(m-n)\pi} \right)\right]\\[10pt]
&=\frac{1}{2}\sum_{m=1}^M\sum_{n=1}^N\bigg[\operatorname{gd}\!\big((m+n)\pi\big)-\operatorname{gd}\!\big((m-n)\pi\big)\bigg]
\end{align}
$$
and the desired integral is $I(5,3)$.

Sorry for the Cleo-style answer but right now I'm busy playing Pokemon Go, so I'll post the complete solution when I'm free. See ya...