Find the Fourier serie

I got this set of functions defined on [-pi, pi]\{0} by

[tex]f_n(x)=\frac{sin((n+1/2)x)}{sin(x/2)}[/tex]

and am asked to find its fourier expansion.

The integral a_n is pretty messy. I've tried using the identity sinAcoB = 0.5(...) but then I'm left with two integrals of the type sin([n±m+1/2]x)/sin(x/2). I've tried pluging numerical values of n and m in "The integrator" and the results are not happy-looking. Type for exemple "Sin[1.5x]/Sin[10x]" at http://integrals.wolfram.com/.

So I was wondering if there was a shortcut to this integral? Maybe linked to the integral bounds? I've found that the (smallest )period of sin([n-m+1/2]x) is [itex]4\pi/(2(n-m)+1)[/itex] while that of sin(x/2) is 4pi. So they are both of period 4pi.

 

I must have made a mistake cuz I end up with the serie being infinite and I didn't use a Taylor expansion.

What I did:

[tex]a_m=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{sin([n+1/2]x)}{sin(x/2)}cos(mx)dx = \frac{1}{\pi}\int_0^{\pi}\mbox{Im}\left\{\frac{e^{i([n-m+1/2]x)}+e^{i([n+m+1/2]x)}}{e^{i(x/2)}}\right\}dx[/tex]

where I use the trig identity sinAcosB = {sin(A-B) + sin(A+B)}/2 before switching to complex exponential notation.

Now set a = m - n and b = m + n. and multiply the integrand by 1 in the form [itex]e^{i(x/2)}e^{-i(x/2)}[/itex], which leads to

[tex]a_m = \frac{1}{\pi}\int_{0}^{\pi}\mbox{Im}\left\{e^{iax}+e^{ibx}\right\}dx[/itex]

Anything wrong with that so far?