Does $\int_{0}^{\infty}{\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 16}?$ How do we prove these two results?
$$\int_{0}^{\infty}{\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 16}\tag1$$
$$\int_{0}^{\infty}{\sin(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 8\pi}\tag2$$
Try an approach to split out the form 
$${\cos(10x^2\pi)\sin(6x^2\pi)\over \sinh^2(2x\pi)}\mathrm dx={1\over 4}{\sin(16x^2\pi)-\sin(4x^2\pi)\over 2\sinh(x\pi)\cosh(x\pi)}\tag3$$
$$={1\over 4}{\sin(16x^2\pi)-4\sin(x^2\pi)\cos(x^2\pi)+8\sin^3(x^2\pi)\cos(x^2\pi)\over 2\sinh(x\pi)\cosh(x\pi)}\tag4$$
 Or we could write $(1)$ as 
$$\int_{0}^{\infty}{({e^{i10x^2\pi}+e^{-i10x^2\pi}})({e^{i6x^2\pi}-e^{-i6x^2\pi}})\over4i \sinh^2(2x\pi)}\mathrm dx\tag5$$
$$\int_{0}^{\infty}{{e^{i16x^2\pi}-e^{-i16x^2\pi}}-({e^{i4x^2\pi}-e^{-i4x^2\pi}})\over4i \sinh^2(2x\pi)}\mathrm dx\tag6$$
$$\int_{0}^{\infty}{\sinh(i16x^2\pi)-\sinh(i4x^2\pi)\over2i \sinh^2(2x\pi)}\mathrm dx\tag7$$
Surely this is not the correct  approach here.
I estimated the closed form using wolfram integrator, not sure it is correct.

I've actually no idea how to proof it, but the question got me interested and I tried to do a few things. You could try to break them all down to exponentials. This is what I wrote on MathType for formula (1):

I would suggest to substitute the $e^{2\pi x}$ into a value $a$ and eventually you should get four integrals with something like 
$ \left(\frac{ca^2}{a^2-1}\right) $. Evaluating this integral should give a logaritm and subsituting the value back could give the pretty number you were looking for. Good luck!
