integration - Evaluating $int_0^pi frac{cos{ntheta}}{1 -2rcos{theta}+r^2} dtheta$

How do I go about evaluating the following integral:

$$\int_0^\pi \frac{\cos{n\theta}}{1 -2r\cos{\theta}+r^2} d\theta$$

for $n \in \mathbb{N}$, and $r \in (0,1)$?

My tack has been to rewrite, using the exponential form of $\cos$, as:

$$\int_0^\pi \frac{e^{i\theta} + e^{-i\theta}}{2(1 - re^{i\theta})(1-re^{-i\theta})} d\theta$$

and letting $z = e^{i\theta}$ and $d\theta = \frac{1}{iz}dz$, we get

$$\frac{1}{2i}\int_{|z|=1} \frac{z^n + \frac{1}{z}^n}{(1 - rz)(z-r)} d\theta$$

This would have a singularity when $z= e^{i\theta} = r$ or when $zr = re^{i \theta} = 1$. Since $r \in (0,1)$ neither of these can occur. So we just need to integrate the above, but I can't see any way about this?

Is my approach valid or along the right lines, and how can I proceed?

Answer

The contour integration way:

Residue theorem reveals within a second that

$$\int_0^\pi \frac{\cos{n\theta}}{1 -2r\cos{\theta}+r^2} d\theta=\pi\frac{r^n}{1-r^2}.$$

Note: To make everything simpler (compared with what you tried in your post) it is enough to use in the numerator $e^{i n\theta}$.

The real method way:

Exploit carefully the well-known series result (which can be proved by real methods)

$$\sum_{n=1}^{\infty} p^n \sin(n x)=\frac{p\sin(x)}{1-2 p \cos(x)+p^2}, |p|<1$$