Note first that the integrand is even and can be extended to the entire real line. Consider
$$f(z) = \frac{e^{i b z}}{\sinh{a z}}$$
$b>0$, in the upper half complex plane. Then
$$\oint_C dz \: \frac{e^{i b z}}{\sinh{a z}}$$
is equal to $i 2 \pi$ times the sum of the residues within $C$, which we'll take to be a semicircle of large radius in the upper half plane indented at the origin to include the pole at $z=0$. The integral about the semicircular arc goes to zero as this radius goes to infinity because of Jordan's Lemma. Then the integral of $f$ along the real line is the sum of the residues within $C$.
$\sinh{a z}$ has poles at $z=i n \pi/a$, $n \ge 0$. The residue term here is a little tricky, but may be shown to be equal to
$$\text{Res}_{z=i n \pi/a} f(z) = \frac{(-1)^n}{a} e^{-n \pi b/a}$$
The integral along the real line is then the sum of all the residues within $C$:
$$\int_{-\infty}^{\infty} dx \: \frac{e^{i b x}}{\sinh{a x}} =\frac{i 2 \pi}{a} \sum_{n=0}^{\infty} (-1)^n e^{-n \pi b/a} = \frac{i 2 \pi}{a} \frac{1}{1+e^{-\pi b/a}}$$
For the case $b<0$, we use a contour in the lower half plane but indented to exclude the pole at $z=0$ (which was already included before); the result is:
$$\int_{-\infty}^{\infty} dx \: \frac{e^{-i b x}}{\sinh{a x}} =\frac{i 2 \pi}{a} \sum_{n=1}^{\infty} (-1)^n e^{-n \pi b/a} = -\frac{i 2 \pi}{a} \frac{1}{1+e^{\pi b/a}}$$
Then we take $1/2$ the sum to get the integral over the positive reals:
$$\int_{0}^{\infty} dx \: \frac{\sin{b x}}{\sinh{a x}} = \frac{\pi}{2 a} \tanh{\left(\pi \frac{b}{2 a}\right)}$$
By using the Maclaurin series of $\exp(x)$ and $\cos(x)$ both integrals boil down to the computation of
$$ \int_{0}^{\pi/2}\frac{\left(\cos x\right)^n}{a^2 \sin^2(x)+b^2\cos^2(x)}\,dx=\int_{0}^{\pi/2}\frac{\left(\sin x\right)^n}{a^2 \sin^2(x)+b^2\cos^2(x)}\,dx $$
which can be tackled thrugh the tangent half-angle substitution and the residue theorem.
Best Answer
First, use the identity $\sin (x-y)=\sin x \cos y - \cos x \sin y$, and the sum rule to get $$\int_a^b\left[\int_a^y\frac{\sin x\cos y}{xy}\mathrm{d}x-\int_a^y\frac{\cos x\sin y}{xy} \mathrm{d}x \right]\mathrm{d}y$$ Pull out the constants and you get $$\int_a^b\left[ \frac{\cos{y}}{y}\int_a^y\frac{\sin x}{x}\mathrm{d}x-\frac{\sin y}{y}\int_a^y\frac{\cos x}{x}\mathrm{d}x \right]\mathrm{d}y$$
This simplifies to $$\int_a^b\left[\frac{\cos y}{y}\Bigr[\mathrm{Si}(x)\Bigr]_a^y-\frac{\sin y}{y}\Bigr[\mathrm{Ci}(x)\Bigr]_a^y\right]\mathrm{d}y$$
$$\int_a^b\left[\frac{\cos y}{y}\Bigr(\mathrm{Si}(y)-\mathrm{Si}(a)\Bigr)-\frac{\sin y}{y}\Bigr(\mathrm{Ci}(y)-\mathrm{Ci}(a)\Bigr)\right]\mathrm{d}y$$
$$\int_a^b\left[\frac{\mathrm{Si}(y)\cos y}{y}-\frac{\mathrm{Si}(a)\cos y}{y}-\frac{\mathrm{Ci}(y)\sin y}{y}+\frac{\mathrm{Ci}(a)\sin y}{y}\right]\mathrm{d}y$$
$$\int_a^b\frac{\mathrm{Si}(y)\cos y}{y}\mathrm{d}y-\int_a^b\frac{\mathrm{Si}(a)\cos y}{y}\mathrm{d}y-\int_a^b\frac{\mathrm{Ci}(y)\sin y}{y}\mathrm{d}y+\int_a^b\frac{\mathrm{Ci}(a)\sin y}{y}\mathrm{d}y$$
$\mathrm{Si}(a)$ and $\mathrm{Ci}(a)$ are constants, so they come out of their respective integrals, which further simplify. $$\int_a^b\frac{\mathrm{Si}(y)\cos y}{y}\mathrm{d}y-\mathrm{Si}(a)\int_a^b\frac{\cos y}{y}\mathrm{d}y-\int_a^b\frac{\mathrm{Ci}(y)\sin y}{y}\mathrm{d}y+\mathrm{Ci}(a)\int_a^b\frac{\sin y}{y}\mathrm{d}y$$
$$\int_a^b\frac{\mathrm{Si}(y)\cos y}{y}\mathrm{d}y-\mathrm{Si}(a)\Bigr[\mathrm{Ci}(y)\Bigr]_a^b-\int_a^b\frac{\mathrm{Ci}(y)\sin y}{y}\mathrm{d}y+\mathrm{Ci}(a)\Bigr[\mathrm{Si}(y)\Bigr]_a^b$$
$$\mathrm{Si}(a)\mathrm{Ci}(a)-\mathrm{Si}(a)\mathrm{Ci}(b)+\mathrm{Si}(b)\mathrm{Ci}(a)-\mathrm{Si}(a)\mathrm{Ci}(a)+\int_a^b\frac{\mathrm{Si}(y)\cos y}{y}\mathrm{d}y-\int_a^b\frac{\mathrm{Ci}(y)\sin y}{y}\mathrm{d}y$$
$$\mathrm{Si}(b)\mathrm{Ci}(a)-\mathrm{Si}(a)\mathrm{Ci}(b)+\int_a^b\frac{\mathrm{Si}(y)\cos y}{y}\mathrm{d}y-\int_a^b\frac{\mathrm{Ci}(y)\sin y}{y}\mathrm{d}y$$
Using integration by parts, we can transform $-\int_a^b\frac{\mathrm{Ci}(y)\sin y}{y}\mathrm{d}y$ into the other integral. $$-\int_a^b\frac{\mathrm{Ci}(y)\sin y}{y}\mathrm{d}y = -\Bigr[\mathrm{Si}(y)\mathrm{Ci}(y)\Bigr]_a^b+\int_a^b\frac{\mathrm{Si}(y)\cos y}{y}\mathrm{d}y$$
Substituting, we get $$\mathrm{Si}(b)\mathrm{Ci}(a)-\mathrm{Si}(a)\mathrm{Ci}(b)+\mathrm{Si}(a)\mathrm{Ci}(a)-\mathrm{Si}(b)\mathrm{Ci}(b)+2\int_a^b\frac{\mathrm{Si}(y)\cos y}{y}\mathrm{d}y$$
$$\big(\mathrm{Si}(a)+\mathrm{Si}(b)\big)\big(\mathrm{Ci}(a)-\mathrm{Ci}(b)\big)+2\int_a^b\frac{\mathrm{Si}(y)\cos y}{y}$$
Now, to the best of my knowledge, that integral has no elementary solution. After laboring over it, I finally ran it through Maxima, which uses the Risch Algorithm, and it got nothing as well. However, we can expand the sine integral as a convergent infinite series (source: [Wikipedia][1]), and then solve the integral from there. $$\mathrm{Si}(y)=\sum_{n=0}^\infty\frac{(-1)^ny^{2n+1}}{(2n+1)(2n+1)!}$$ Substituting this in, we get $$2\int_a^b\frac{\mathrm{Si}(y)\cos y}{y}\mathrm{d}y=\sum_{n=0}^\infty\frac{2(-1)^n}{(2n+1)(2n+1)!}\int_a^b\frac{y^{2n-1}\cos y}{y}\mathrm{d}y$$ $$\sum_{n=0}^\infty\frac{2(-1)^n}{(2n+1)(2n+1)!}\int_a^b y^{2n}\cos y\,\mathrm{d}y$$ That integral is solvable, and can be found to be $$\int_a^b y^{2n}\cos y\,\mathrm{d}y=\Bigr[-\frac{1}{2}i^{2n+1}\big[\Gamma(2n+1,-ix)+(-1)^{2n}\Gamma(2n+1,ix)\big]\Bigr]_a^b$$ Plugging this in we get the final answer. $$-\sum_{n=0}^\infty\frac{i^{4n+1}}{(2n+1)(2n+1)!}\big[\Gamma(2n+1,-ib)+(-1)^{2n}\Gamma(2n+1,ib)-\Gamma(2n+1,-ia)-(-1)^{2n}\Gamma(2n+1,ia)\big]$$ $$\big(\mathrm{Si}(a)+\mathrm{Si}(b)\big)\big(\mathrm{Ci}(a)-\mathrm{Ci}(b)\big)-\sum_{n=0}^\infty\frac{i^{4n+1}}{(2n+1)(2n+1)!}\big[\Gamma(2n+1,-ib)+(-1)^{2n}\Gamma(2n+1,ib)-\Gamma(2n+1,-ia)-(-1)^{2n}\Gamma(2n+1,ia)\big]$$ I have it worked out without the incomplete Gamma function as a double sum. If you want me to write out that solution as well, leave a comment.
EDIT: Here's the solution involving infinite sums. Now, because n is discrete, I can describe the integral as an infinite sum. I solved the integrals for the cases $n = 0$, $n=1$, and $n=2$, and constructed the rule $$\int_a^b\frac{y^{2n-1}\cos y}{y}\mathrm{d}y=\left[\sum_{q=0}^{2n}(-1)^{q+1}\frac{2n!}{q!}\frac{\mathrm{d}^{q}\sin x}{\mathrm{d}x^{q}}x^q\right]_a^b$$ $$\sum_{q=0}^{2n}(-1)^{q+1}\frac{2n!}{q!}\frac{\mathrm{d}^{q}\sin b}{\mathrm{d}x^{q}}b^q-\sum_{q=0}^{2n}(-1)^{q+1}\frac{2n!}{q!}\frac{\mathrm{d}^{q}\sin a}{\mathrm{d}x^{q}}a^q$$ (I tested this for the case $n=3$, and it works fine). So, substituting this into the equation, we get $$\sum_{n=0}^\infty\left[\frac{2(-1)^n}{(2n+1)(2n+1)!}\left[\sum_{q=0}^{2n}\left[(-1)^{q+1}\frac{2n!}{q!}\frac{\mathrm{d}^{q}\sin x}{\mathrm{d}x^{q}}x^q\right]\right]_a^b\right]$$ $$\big(\mathrm{Si}(a)+\mathrm{Si}(b)\big)\big(\mathrm{Ci}(a)-\mathrm{Ci}(b)\big)+2\sum_{n=0}^\infty\left[\frac{2(-1)^n}{(2n+1)(2n+1)!}\left[\sum_{q=0}^{2n}\left[(-1)^{q+1}\frac{2n!}{q!}\frac{\mathrm{d}^{q}\sin x}{\mathrm{d}x^{q}}x^q\right]\right]_a^b\right]$$