Integration by parts (just to simplify the evaluation) gives $$\int_0^\infty\frac{\sinh ax\sinh bx}{(\cosh ax+\cos t)^2}\,dx=\frac{b}{a}\int_0^\infty\frac{\cosh bx\,dx}{\cosh ax+\cos t}.$$ This can be evaluated using the residue theorem. Consider $I_R=\displaystyle\int_{C_R}\frac{e^{bz}\,dz}{\cosh az-\cos t}$ where $C_R$ is the rectangular contour with vertices at $\pm R\pm i\pi/a$; the integrand has simple poles at $z=\pm it/a$, thus $$I_R=2\pi i\left(\operatorname*{Res}_{z=it/a}+\operatorname*{Res}_{z=-it/a}\right)\frac{e^{bz}}{\cosh az-\cos t}=\frac{4\pi i}{a}\frac{\sin(bt/a)}{\sin t},$$ and on the other hand, $\lim\limits_{R\to\infty}I_R$ is equal to $$\int_{-\infty}^\infty\frac{e^{b(x-i\pi/a)}\,dx}{-\cosh ax-\cos t}+\int_\infty^{-\infty}\frac{e^{b(x+i\pi/a)}\,dx}{-\cosh ax-\cos t}=4i\sin\frac{b\pi}{a}\int_0^\infty\frac{\cosh bx\,dx}{\cosh ax+\cos t}.$$
The formula of the question follows easily from Ramanujan's generalization of Frullani's integral. See 'The Quarterly Reports of S. Ramanujan,' American Mathematical Monthly, Vol. 90, #8 Oct 1983, p 505-516. I won't give the conditions, but state it to establish the notation. Let
$$ f(x)-f(\infty)=\sum_{k=0}^\infty u(k)(-x)^k/k! \quad,\quad g(x)-g(\infty)=\sum_{k=0}^\infty v(k)(-x)^k/k!$$
$$ f(0)=g(0) \quad,\quad f(\infty)=g(\infty) $$
Then
$$\int_0^\infty \frac{dx}{x} \big(f(ax) - g(bx) \big)= \big(f(0)-f(\infty) \big)\Big( \log{(b/a)} +
\frac{d}{ds} \log{\Big(\frac{v(s)}{u(s)}\Big)}\Big|_{s=0} \Big) $$
For the OP's case, a=b=1, $f(x)=2qe^{-x} \implies u(k)=2q, \ f(0)=2q, f(\infty)=0.$
We will eventually show
$$ (1) \quad g(x)=\frac{\sinh(q \ x)}{\sinh(x/2)} = -\sum_{n=0}^\infty \frac{(-x)^n}{n!} \ \cos(\pi \ n) \big( \zeta(-n, 1/2+q) - \zeta(-n, 1/2-q) \big)$$
where $\zeta(s,a)$ is the Hurwitz zeta function.
Given (1), it is easy to see that
$$ \frac{d}{ds} \log{v(s)} \big|_{s=0} = \frac{v'(0)}{v(0)} = -\frac{ \zeta'(0, 1/2+q)-\zeta'(0, 1/2-q)}{ \zeta(0, 1/2+q)-\zeta(0, 1/2-q) }$$
However, it is known that
$$\zeta'(0,a)=\log(\Gamma(a)/\sqrt{2\pi}) \text{ and } \zeta(0,a)=-B_1(a)=1/2-a $$ where in the last formula the Hurwitz zeta has been connected to the Bernoulli polynomial by the formula
$$ (2) \quad \zeta(-n,a) = -\frac{B_{n+1}(a)}{n+1}. $$
Doing the rest of the algebra results in the expression
$$ (3) \quad \int_0^\infty \Big(2qe^{-x} - \frac{ \sinh(q \ x)}{\sinh{x/2} } \Big) \frac{dx}{x} = \log{\Gamma(1/2+q)} - \log{\Gamma(1/2-q)}. $$ To get it in the form of the OP's request, use the gamma-function reflection formula
$$ \Gamma(1/2-q)\Gamma(1/2+q) = \frac{\pi}{\cos{\pi q } } .$$
Now, to prove (1):
$$ \frac{\sinh(q \ x)}{\sinh(x/2)} = \frac{e^{qx} - q^{-qx}}{e^{x/2}-e^{-x/2}} =
\frac{1}{x}\Big( \frac{x}{e^x-1} \exp(x(1/2+q))+\frac{x}{e^x-1} \exp(x(1/2-q)) \Big)
$$
Use the well-known generating function for Bernoulli polynomials,
$$ \frac{\sinh(q \ x)}{\sinh(x/2)} =\frac{1}{x}\sum_{n=0}^\infty \frac{x^n}{n!} \Big( B_n(1/2+q) - B_n(1/2-q) \Big) =\sum_{n=0}^\infty \frac{x^n}{n!(n+1)} \Big( B_{n+1}(1/2+q) - B_{n+1}(1/2-q) \Big) $$
where in the second step we have re-indexed because $B_0(x)=1$ and the first term is thus zero. Then use (2) in the last formula to complete the proof of (1).
Best Answer
For $z\in\mathbb{C}$, $|z|<1$, $0<u<1$, we have (a known "hypergeometric" representation) $$I(z):=\int_0^1\frac{x^{u-1}(1-x)^{-u}}{1+zx}\,dx=\sum_{n=0}^{\infty}(-z)^n\mathrm{B}(n+u,1-u)\\=\frac{\pi}{\sin u\pi}\sum_{n=0}^{\infty}\binom{u+n-1}{n}(-z)^n=\frac{\pi}{\sin u\pi}(1+z)^{-u}$$ (with the principal branch of $(\cdot)^{-u}$). Your integral is equal to $\dfrac{e^{it}I(ae^{it})-e^{-it}I(ae^{-it})}{2i\sin t}$.