Fourier Coefficients for $\frac{3}{5-4\cos{2\theta}}$ I want to compute the (even) fourier coefficients for 
 $\dfrac 3 {5-4\cos 2\theta}$ on the interval $[0, 2\pi]$.
Namely, I want to compute the integral:
$$b_n = \int_0^{2\pi} \cos(n\theta) \frac 3 {5-4\cos 2\theta } \frac{d\theta}{2\pi}$$
Integrating this in mathematica, it seems like $b_n = 0$ if $n$ is odd and $b_n = \left(\frac 1 2 \right)^{n/2}$ for $n$ even. 
 A: Hint: Observe
\begin{align}
\frac{3}{5-4\cos 2\theta} = \frac{3/5}{1-\frac{4}{5}\cos 2\theta} = \frac{3}{5}\sum^\infty_{n=0} \left(\frac{4}{5}\right)^n \cos^n 2\theta
\end{align}
and
\begin{align}
\cos^n 2\theta = \left(\frac{e^{2i\theta}+e^{-2i\theta}}{2}\right)^n = \frac{1}{2^n}\sum^n_{k=0}\binom{n}{k} e^{2ik\theta}e^{-2i(n-k)\theta}.
\end{align}
Thus
\begin{align}
\int^{2\pi}_0 e^{2ik\theta}e^{-2i(n-k)\theta} \frac{e^{in\theta}+e^{-in\theta}}{2}\ dx = \text{ something }.
\end{align}
Of course, you could have just consider the integral
\begin{align}
\int^{2\pi}_0 \cos(n\theta) \cos^{n} 2\theta\ \frac{d\theta}{2\pi}.
\end{align}
A: Complex analysis approach:
Let $$f(z)=\frac{3}{5-2(z^2+z^{-2})}=\frac{-3z^2}{2z^4-5z^2+2}=\frac{1}{2z^2-1}-\frac{2}{z^2-2}$$ Then $f(e^{i\theta})=\frac{3}{5-4\cos 2\theta}$.
Now, for $|z|<\sqrt{2}$, we have:
$$\frac{-2}{z^2-2}=\frac{1}{1-z^2/2}=\sum_{k=0}^{\infty} \frac{1}{2^k}z^{2k}$$
and for $|z|>\frac{1}{\sqrt{2}}$ you have:
$$\frac{1}{2z^2-1}=\frac{1}{2z^2}\frac{1}{1-\frac{1}{2z^2}}=\sum_{k=1}^{\infty}\frac{1}{2^k}z^{-2k}$$
Since $z=e^{i\theta}$ is in this range, this means that:
$$\frac{3}{5-4\cos 2\theta} = \sum_{k=\infty}^\infty \frac{1}{2^{|k|}}e^{i2k\theta}$$
Combining $k$ and $-k$ terms, this gives:
$$\frac{3}{5-4\cos2\theta} = 1 + \sum_{k=1}^{\infty} \frac{1}{2^{k-1}}\cos 2k\theta$$
I might have missed something in this computation, but this is close to your answer - the coefficients are $0$ if $n$ is odd and $\frac{1}{2^{k-1}}$ if $n=2k$.
