Integral $\int_{-\infty}^{+\infty} \operatorname{sech}^2(x) \cos (2x) \, dx=\frac{2\pi}{\operatorname{sinh}(\pi)}$ Show that $$\int_{-\infty}^{+\infty} \operatorname{sech}^2(x) \cos (2x) \, dx=\frac{2\pi}{\operatorname{sinh}(\pi)}$$
I could not start because the integrand does not have any singularities. I know the whole idea of doing such problem by method of complex integration but in this particular problem I stuck.
I expanded the integrand in Taylor series and saw the sum  converges( therefore the integration is possible. I think I am missing little idea.
Any hints would really be appreciated!

Let $I$ denote the integral. Then 
$$ I = \int_{-\infty}^{\infty} \frac{e^{2ix}}{\cosh^{2}x} \, dx. $$
Now considering a CCW contour $C_{R}$ traversing the rectangle formed by four vertices $\pm R$ and $\pm R + \pi i$, we find that
$$ \int_{C_{R}} \frac{e^{2iz}}{\cosh^{2}z} \, dz = 2\pi i \operatorname{Res}\limits_{z=i\pi/2} \frac{e^{2iz}}{\cosh^{2}z} = 4\pi e^{-\pi}. $$
On the other hand, taking $R\to\infty$ it is not hard to show that the LHS converges to $ (1 - e^{-2\pi})I$. (Here, the contribution $I$ follows from the bottom side of $C_{R}$ and the contribution $-e^{-2\pi}I$ follows from the top side of $C_{R}$.) This proves that
$$ I = \frac{4\pi e^{-\pi}}{1 - e^{-2\pi}} = \frac{2\pi}{\sinh \pi}. $$
