Hints on evaluating this $\frac{3}{2\pi}\int_0^{2\pi}\frac{e^{-ikx}}{5 - 4\cos(x)} \mathrm dx$? I have the following integral I'm trying to solve:
$$\frac{3}{2\pi}\int_0^{2\pi}\frac{e^{-ikx}}{5 - 4\cos(x)} \mathrm dx, \quad k \in \mathbb{Z}.$$
I've tried writing the exponential in terms of sines and cosines and then using the usual rational trig function substitutions, i.e. rewriting the integrand as
$$\frac{\cos(kx)}{5 - 4\cos(x)} - \frac{i\sin(kx)}{5 - 4\cos(x)}$$
and then using the substitution $t = \tan{x/2}$, but this doesn't result in anything useful that I could come up because of the different fequencies in the trig functions.  I also tried writing the cosine in terms of complex exponentials, but didn't end up with anything useful either.
If the only solution involves residue calculus then please let me know, as I'm not familiar with that subject so I haven't tried anything like that.  Thanks!
UPDATE: As was suggested I wrote the $\frac{1}{5 - 4\cos(x)}$ term as a geometric series $\frac{1}{5}\sum\limits_{n = 0}^\infty (\frac{4}{5}\cos(x))^n$, so now I am trying to evaluate
$$\int_0^{2\pi} \cos^n(x) e^{-ikx} dx,$$
but am stuck here again.  I am trying to do this without using the residue theorem, but if that's the only way then so be it!
 A: Alternatively you could consider using Residue Theorem by writing $\cos(x)=\frac12(e^{ix}+e^{-ix})$ and substitute into the integrand. By letting $e^{ix}=z$ as a new variable you could transform the integral to a complex integral around the unit circle with argument from $0$ to $2\pi$. Check any complex methods textbook and apply the formula and that should give you the answer.
A: For $k\ge 1$ put $z=e^{ix}$ to obtain
$$\frac{3}{2\pi} \int_{|z|=1} \frac{1/z^k}{5-2z-2/z} \frac{1}{iz} dz
= 3\times \frac{1}{2\pi i} \int_{|z|=1} \frac{1/z^k}{5z-2z^2-2} dz
\\= 3\times \frac{1}{2\pi i} \int_{|z|=1} \frac{1/z^k}{(1-2z)(z-2)} dz
= -\frac{3}{2}\times \frac{1}{2\pi i} \int_{|z|=1} \frac{1/z^k}{(z-1/2)(z-2)} dz.$$
There are two poles inside the unit circle, one  at $z=1/2$
with residue $2^k\times 1/(-3/2).$ 
For the residue at the other pole which is at $z=0$ observe that
$$\frac{1}{(z-1/2)(z-2)} = \frac{4}{3}\frac{1}{1-2z} - \frac{1}{3}\frac{1}{1-z/2}.$$
This gives the residue $$\frac{4}{3} 2^{k-1} -\frac{1}{3} \frac{1}{2^{k-1}}
= \frac{2}{3} 2^k -\frac{2}{3} \frac{1}{2^k}.$$
The first term cancels with the residue at $z=1/2$ and the second term is the only contribution for a final answer of
$$-\frac{3}{2} \times \left( -\frac{2}{3} \frac{1}{2^k} \right)= \frac{1}{2^k}.$$
