How to prove
$$\int_0^{\pi/2} \frac{\cos ((1-a) x)\cos ^{1-a}(x)}{ (\cosh (2 b)-\cos (2 x))} \, dx=\frac{\pi e^{(a-1)b}}{4\sinh (b) \cosh ^a(b)}$$
So far I've got no idea of tackling it (my intuition is on contour integration). I'd like you to give some suggestions. Thanks in advance!
Update: I found an alternate proof for @pisco's general formula. By using Fourier expansion
$$\frac{\sinh (2 b)}{\cosh (2 b)-\cos (2 x)}=2 \sum _{k=1}^{\infty } e^{-2kb} \cos (2 k x)+1$$
and trigonometric identities $\cos*\cos\to \cos+\cos$ , one have
$$\small I=\int_0^{\pi /2} {\frac{{{{(\cos x)}^a}\cos cx}}{{\cosh 2b – \cos 2x}}dx} =\frac{\sum _{k=1}^{\infty } \exp (-2 b k) (f(a+1,c+2 k)+f(a+1,c-2 k))+f(a+1,c)}{\sinh (2 b)}$$
Where $f$ denotes the classic Cauchy integral ($\Re v>0$)
$$f(v,a)=\int_0^{\frac{\pi }{2}} \cos (a x) \cos ^{v-1}(x) \, dx=\frac{\pi }{2^v v B\left(\frac{1}{2} (a+v+1),\frac{1}{2} (-a+v+1)\right)}$$
Performing the summation using definition of hypergeometric functions one have
$$\small I=\frac{\pi 2^{-a-1} \text{csch}(2 b) \left(e^{-2 b} \left(\frac{\, _2F_1\left(1,\frac{1}{2} (-a-c+2);\frac{1}{2} (a-c+4);-e^{-2 b}\right)}{B\left(\frac{a+c}{2},\frac{1}{2} (a-c+4)\right)}+\frac{\, _2F_1\left(1,\frac{1}{2} (-a+c+2);\frac{1}{2} (a+c+4);-e^{-2 b}\right)}{B\left(\frac{1}{2} (a+c+4),\frac{a-c}{2}\right)}\right)+\frac{1}{B\left(\frac{1}{2} (a+c+2),\frac{1}{2} (a-c+2)\right)}\right)}{a+1}$$
Which, after simplifications, should agree with @pisco's result (it passed my numeric verification so I won't simplify it further).
Best Answer
This integral, as well a generalization: $$\tag{*}\int_0^{\pi /2} {\frac{{{{(\cos x)}^a}\cos cx}}{{\cosh 2b - \cos 2x}}dx} \\ = \frac{\pi \Gamma (a+1) \left[\, _2F_1\left(1,-\frac{a+c}{2};\frac{a-c}{2}+1;-e^{-2 b}\right)+\, _2F_1\left(1,\frac{c-a}{2};\frac{a+c}{2}+1;-e^{-2 b}\right)-1\right]}{2^{a+1}\sinh(2b)\Gamma \left(\frac{a-c}{2}+1\right) \Gamma \left(\frac{a+c}{2}+1\right)}$$ are easy consequences of Fourier transform techniques, with $_2F_1$ the hypergeometric function.
Recall (see 1, 2): $$\begin{aligned}\int_{ - \pi /2}^{\pi /2} {{{(\cos x)}^a}{e^{ibx}}dx} &= \frac{{\pi \Gamma (a + 1)}}{{{2^a}\Gamma (1 + \frac{{a + b}}{2})\Gamma (1 + \frac{{a - b}}{2})}} \qquad &\Re(a)>-1, b\in \mathbb{C}\\ \int_{ - \infty }^\infty {\frac{{\sinh dx}}{{\sinh \pi x}}{e^{iux}}dx} &= \frac{{\sin d}}{{\cosh u + \cos d}} \qquad &|\Re (d)|<\pi, u\in \mathbb{R}\end{aligned}$$
For the Fourier transform $\hat{f}(\xi) = \int_\mathbb{R} f(x)e^{-2\pi i x \xi} dx$, apply the Plancherel formula in the following form $$\int_{-\infty}^\infty f(x)\hat{g}(x) dx = \int_{-\infty}^\infty \hat{f}(x)g(-x) dx$$ to ($\chi_A$ represents the characteristic function of set $A$) $$f(x) = (\cos 2\pi x)^a e^{-2\pi i c x}\chi_{(-1/4,1/4)}(x)\qquad g(x) = \sinh \frac{dx}{2b}\text{csch} \frac{\pi x}{2b}$$ with $a>-1, b>0, c\in \mathbb{R}, -\pi<\Re(d)<\pi$ produces $$\small\frac{1}{{2\pi }}\int_{ - \pi /2}^{\pi /2} {{{(\cos x)}^a}{e^{ - icx}}\frac{{2b\sin d}}{{\cosh 4\pi bx + \cos d}}dx} = \frac{{\Gamma (a + 1)}}{{{2^{a + 1}}}}\int_{ - \infty }^\infty {\frac{1}{{\Gamma (1 + \frac{{a + c + x}}{2})\Gamma (1 + \frac{{a - c - x}}{2})}}\frac{{\sinh (dx/2b)}}{{\sinh (\pi x/2b)}}dx} $$ take $a=c$ and $d\mapsto 2id$ (so now $|\Im(d)|<\pi/2$), $$\tag{1}\small \frac{1}{{2\pi }}\int_{ - \pi /2}^{\pi /2} {{{(\cos x)}^a}{e^{ - iax}}\frac{{2\sinh 2d}}{{\cosh 2bx + \cosh 2d}}dx} = \frac{{\Gamma (a + 1)}}{{{2^a}}}\int_{ - \infty }^\infty {\frac{1}{{\Gamma (1 + a + bx)\Gamma (1 - bx)}}\frac{{\sin (2dx)}}{{\sinh \pi x}}dx} $$
Assuming $d>0$ now, I claim in the above expression, we can take $b=i$. Due to Stirling formula, $$|\Gamma (1 + a + bx)\Gamma (1 - bx)|^{-1} \sim C e^{\pi|\Im (b)|x}|x|^{-a-1} \qquad x\in \mathbb{R}, x\to \pm \infty$$ hence RHS of $(1)$ remains absolute convergent when $|\Im (b)|<1$. Since $a>-1$, continuity shows that taking $b=i$ is legitimate. So we arrive at $$\frac{1}{{2\pi }}\int_{ - \pi /2}^{\pi /2} {{{(\cos x)}^a}{e^{ - iax}}\frac{{2\sinh 2d}}{{\cos 2x + \cosh 2d}}dx} = \frac{{\Gamma (a + 1)}}{{{2^a}}}\frac{i}{\pi }\int_{ - \infty }^\infty {\frac{{\Gamma (ix)}}{{\Gamma (1 + a + ix)}}\sin (2dx)dx} $$
Formula $(2)$ below implies the last integral is $ \frac{{-\pi i}}{{\Gamma (1 + a)}}{(1 - {e^{ - 2d}})^a}$, so $$\int_{ - \pi /2}^{\pi /2} {\frac{{{{(\cos x)}^a}\cos ax}}{{\cos 2x + \cosh 2d}}dx} = {2^{ - a}}\frac{{\pi {{(1 - {e^{ - 2d}})}^a}}}{{\sinh 2d}}\qquad a>-1, d>0$$ OP's formula is obtained after $d\mapsto d+i\pi/2$ (this is also feasible). If we hadn't take $a=c$ before $(1)$, then we will obtain $(*)$.
I claim that $$\tag{2}\int_{ - \infty }^\infty {\frac{{\Gamma (ix)}}{{\Gamma (ix + a)}}{\sin bx}dx} = - \pi i\frac{{{{(1 - {e^{ - b}})}^{a - 1}}}}{{\Gamma (a)}} \qquad a,b>0$$
To see this, first consider $$\tag{2a}\int_{ - \infty }^\infty {\frac{{\Gamma (ix)}}{{\Gamma (ix + a)}}{e^{ibx}}dx} $$ Integrate it using a semicircle in upper half plane, with an indention below $0$, Jordan's lemma implies integral around big circle tends to $0$, summing residues (at $x=ni,n\geq 0$) using binomial theorem (see also), we obtain $\frac{{2\pi }}{{\Gamma (a)}}{(1 - {e^{ - b}})^{a - 1}}$. On the other hand, integrate $$\tag{2b}\int_{ - \infty }^\infty {\frac{{\Gamma (ix)}}{{\Gamma (ix + a)}}{e^{ - ibx}}dx} $$ with semicircle in lower half plane, with an indention below $0$, Jordan's lemma again implies integral around big circle tends to $0$, but now the integrand has no pole within the contour, so the integral is $0$. Subtracting $(2a)$ and $(2b)$ shows $(2)$.