How to find $\int_0^{\pi}\frac{\sin n\theta}{\cos\theta-\cos\alpha}d\theta$

I was doing some work in physics and I came up with a definite integral. I tried everything I could but couldn't solve the integral. The integral is $$\int_0^\pi {\sin\left(n\theta\right)\over \cos\left(\theta\right) - \cos\left(\alpha\right)}\,{\rm d}\theta\,, \qquad\qquad n\ \in\ {\mathbb Z}\,,\qquad 0\ \leq\ \alpha\ \leq\ 2\pi$$

There is singularity at $\alpha=\theta$ which increases its difficulty.

I tried complex analysis but couldn't solve it.please help me with method,also provide an answer with proof if you like, I would appreciate it.

• @IvoTerek Everyone knows it's $d\theta$. But I agree he should tell us anyway. Jun 29, 2014 at 6:52
• Sorry guys,I forgot to write $d\theta$ :-) Jun 29, 2014 at 6:54
• I even erased my early comment, I read the question quite fast and didn't saw that he told us $\alpha$ was constant. Being $\mathrm{d}\theta$ was the only way. Anyway I edited it so that people don't do the same as me now haha. No stress. Jun 29, 2014 at 6:56
• Here is the same problem. Jun 29, 2014 at 9:17
• @nbubis Sorry I didn't know that,I will keep it in mind in my future posts :-) Jul 1, 2014 at 1:32

Thanks to complex analysis, it is rather easy to obtain $$\int_0^{\pi}{\cos\left(n\theta\right)\over \cos\left(\theta\right) - \cos\left(\alpha\right)}\,{\rm d}\theta=\frac{\pi cos(n\alpha)}{sin(\alpha)}$$ By the way, this result is also obtained in attachment, but with a method much more complicated than usual. In fact, this complicated method is proposed for the much more difficult integral :
$$\int_0^{\pi}{\sin\left(n\theta\right)\over \cos\left(\theta\right) - \cos\left(\alpha\right)}\,{\rm d}\theta\$$ The closed form obtained involves the Incomplete Beta function in the complex range. Nevertheless, one of the parameters is nul, which makes think that further simplification might be possible (may be involving polylogarithms)
Several numerical checking were done. Of course, for the numerical computation of the integral (noted $I$ in attachement), the Cauchy principal value is considered.