Is there a formula for $\int_0^\pi\frac{\sin(nx)}{\cos(\theta)-cos(x)}dx$? I'm working on a school project about the aerodynamics of an helicopter blade. I'm trying to adapt my aerodynamics class ( centered around planes wings mostly ) to this case.
In our class, we used what we called the integrals of Glauert and more specifically, this integral :
$$
\int_{0}^{\pi}\frac{\cos\left(nx\right)}
{\cos\left(\theta\right) - \cos\left(x\right)}\,{\rm d}x = -\pi\,\frac{\sin\left(n\theta\right)}{\sin\left(\theta\right)}
$$
In the helicopter case, I need to evaluate a similar integral but with $\sin\left(nx\right)$ instead of $\cos\left(nx\right)$.
I found a beautiful proof of the first result in the book Inside Interesting Integrals by Paul Nahin ( page $60$ ) which uses recursivity. I tried to adapt it to the other integral using the formula
$$\sin\left(\left(n + 1\right)x\right) +
\sin\left(\left(n - 1\right)x\right)
= 2\sin\left(nx\right)\cos\left(x\right)$$
$$
\mbox{Calling}\quad
I_{n} =
\int_0^\pi\frac{\sin\left(nx\right)}{\cos\left(\theta\right) - \cos\left(x\right)}\,{\rm d}x,
$$
I find that
$$
I_{n+1} - 2\cos\left(\theta\right)I_{n} +
I_{n - 1} = \int_{0}^{\pi}2\sin\left(nx\right)\,{\rm d}x
$$ but I then have a problem to solve the recursive equation be the result is $0$ or $\frac{4}{n}$ depending on the parity of n. I decoupled the system to get two recursive equations for the different parities :

*

*For $a_n=I_{2n}$, I get $a_{n+1}-2\cos(2\theta)a_n+a_{n-1}=8\cos\theta(\frac{1}{2n+1}+\frac{1}{2n-1}$)

*And for $b_n=I_{2n+1}$, I get $b_n-2\cos(2\theta)b_{n-1}+b_{n-2}=\frac{8}{2n-1}\cos(\theta)$
I entered these formula on Wolfram Alpha and got some really ugly results. I was thus wondering if a more simple formula could exist for my integral.
I also quickly looked at the wikipedia proof which uses residual theorems but I got lost quite fast...
Do you know some other way to evaluate this integral or maybe if it's possible to solve the recursive equations to get to a simpler result than the one from Wolfram ?
Thank you very much !
 A: Probably, there is no a manageable general expression but we can get an exact answer for  moderate values of n.
The idea is to get rid of singularity in the integrand.
So we replace $\theta$ with $i\theta$ ($i$ is the imaginary unit).
Since $\cos(i\theta)=\cosh(\theta)$ we get
$$J_n(\theta)=\int_0^\pi\frac{\sin(nx)}{\cosh(\theta)-\cos(x)}dx$$
Now, $J_1(\theta)=-2\ln(\tanh\frac{\theta}{2})$
To return to the original integral we replace $\theta$ with $i\theta$ again and equate the real parts on both sides of the equal sign:
$$I_1(\theta)=P\;V\int_0^\pi\frac{\sin(x)}{\cos(\theta)-\cos(x)}dx=-2\ln(\tan\frac{\theta}{2})$$
Completely analogously:
$$J_2(\theta)=\int_0^\pi\frac{\sin(2x)}{\cosh(\theta)-\cos(x)}dx$$
$$=2\cosh(\theta)\ln\frac{\cosh(\theta)+1}{\cosh(\theta)-1}-4$$
and $$I_2(\theta)=P\;V\int_0^\pi\frac{\sin(2x)}{\cos(\theta)-\cos(x)}dx$$
$$=4\cos(\theta)\ln(\cot\frac{\theta}{2})-4$$
For $n=3,4,...$ the procedure is the same.
REMARK:
In similar manner we can establish the initial integral.
Namely, consider
$$\int_0^\pi\frac{\cos(nx)}{\cosh(\theta)-\cos(x)}dx=\pi\frac{e^{-n\theta}}{\sinh(\theta)}$$
Replacing here $\theta$ with $i\theta$ ($i$ is the imaginary unit) and equating real parts gives
$$P\;V\int_0^\pi\frac{\cos(nx)}{\cos(\theta)-\cos(x)}dx=-\pi\frac{\sin(n\theta)}{\sin(\theta)}$$
