Just to show that evaluation is possible with complex analysis, I will outline the following solution. I warn, though, that I merely get an alternative ugly expression for this integral. Claude's expression is about as good as we likely will get for this one.
Consider the contour integral
$$\oint_C dz \frac{e^{i \omega z}}{\sinh^2{a z}}$$
where $C$ is the rectangle having vertices $\pm i \pi/(2 a)$ and $R \pm i \pi/(2 a)$, and $a=1/\sqrt{2}$. Note that we will need to deform $C$ at the origin by a semicircle of radius $\epsilon$ to avoid a pole there. Then by respective parametrization of each segment of $C$, the contour integral is
$$\int_0^R dx \frac{e^{i \omega x} e^{\pi \omega/(2 a)}}{-\cosh^2{a x}} + i \int_{-\pi/(2 a)}^{\pi/(2 a)} dy \frac{e^{i \omega R} e^{-\omega y}}{\sinh^2{(R+i y)}}\\ + \int_R^0 dx \frac{e^{i \omega x} e^{-\pi \omega/(2 a)}}{-\cosh^2{a x}} + i \int_{\pi/(2 a)}^{\epsilon} dy \frac{e^{-\omega y}}{-\sin^2{a y}} \\+ i \epsilon \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{e^{i \omega \epsilon e^{i \phi}}}{\sinh^2{(a \epsilon e^{i \phi})}}+ i \int_{-\epsilon}^{-\pi/(2 a)} dy \frac{e^{-\omega y}}{-\sin^2{a y}}$$
As $R \to \infty$, the second integral vanishes. As $\epsilon \to 0$, the fifth integral has a limit
$$\frac{i \epsilon}{a^2} \int_{\pi/2}^{-\pi/2} d\phi \, e^{i \phi} \frac{1+i \epsilon \omega e^{i \phi}}{\epsilon^2 e^{i 2 \phi}} \sim -i \frac{2}{a^2} \frac{1}{\epsilon} +\frac{\pi \omega}{a^2}$$
In this limit, the fourth and sixth integrals, and the $1/\epsilon$ term, combine to form a Cauchy principal value. By Cauchy's theorem, the contour integral is zero because there are no poles of the integrand inside $C$. Therefore,
$$2 \sinh{\frac{\pi \omega}{2 a}} \int_0^{\infty} dx \frac{e^{i \omega x}}{\cosh^2{a x}} = \frac{\pi \omega}{a^2} + i \, PV \int_{-\pi/(2 a)}^{\pi/(2 a)} dy \frac{e^{-\omega y}}{\sin^2{a y}}$$
Again, note that by $PV$, I mean
$$\lim_{\epsilon \to 0} \left [\int_{-\pi/(2 a)}^{-\epsilon} dy \frac{e^{-\omega y}}{\sin^2{a y}} + \int_{\epsilon}^{\pi/(2 a)} dy \frac{e^{-\omega y}}{\sin^2{a y}} - \frac{2}{a^2} \frac{1}{\epsilon}\right ]$$
Taking imaginary parts, I get that
$$2 \sinh{\frac{\pi \omega}{2 a}} \int_0^{\infty} dx \frac{\sin{\omega x}}{\cosh^2{a x}} = \, PV \int_{-\pi/(2 a)}^{\pi/(2 a)} dy \frac{e^{-\omega y}}{\sin^2{a y}}$$
which really doesn't help matters much. The integrand has an antiderivative involving hypergeometric functions; if you went this route, you would have to take the limit defined above and that seems a lot to ask. Ugly all around. Then again, I hope this at least illustrates how complex analysis may be used to derive an alternative expression for the integral.
Note that, when we take real parts, we get a lovely integral for free:
$$\int_0^{\infty} dx \frac{\cos{\omega x}}{\cosh^2{a x}} = \frac{\pi \omega}{2 a^2} \operatorname*{csch}{\frac{\pi \omega}{2 a}}$$