Integration $\int_0^{2\pi} \frac{\cos^2 3\theta d\theta}{5-4\cos2\theta}$ by using residues 
$$\int_0^{2\pi} \frac{\cos^2 3\theta d\theta}{5-4\cos2\theta}$$
By substituting $\cos m\theta$ to $\frac{z^m+z^{-m}}{2}$ and $d\theta$ to $\frac{-i}{z}dz$,I get 
$$\int_0^{2\pi} \frac{\cos^2 3\theta d\theta}{5-4\cos2\theta} =\int_C \frac{i(z^6+1)^2}{4z^5(z^2-2)(2z^2-1)}dz$$
where $C$ is a unit circle.
and I found out that isolated singular points of $\frac{i(z^6+1)^2}{4z^5(z^2-2)(2z^2-1)}$ in $C$ are $$z_1=0$$ $$z_2=1/\sqrt2$$ $$z_3=-1/\sqrt2$$
So only what I have to do is calculating their residues and smmanding, and mutiplying $2\pi$.
But as you can see, their order of pole are so high, so that I can't handle it with handwriting. Would be there any simplified way to get 
$$\frac{3\pi}{8}$$
for the answer?

Write $\cos^2{3 \theta} = (1 +  \cos{6 \theta})/2$.  Evaluate each resulting integral separately.  You can eliminate the pole at $z=0$ by writing $\cos{6 \theta} = \operatorname{Re}{\left (e^{i 6 \theta}\right )} $.  The rest is straightforward.
To illustrate, the integral is
$$\frac12 \int_0^{2 \pi} d\theta \frac1{5 - 4 \cos{2 \theta}} + \frac12 \operatorname{Re}{ \int_0^{2 \pi} d\theta \frac{e^{i 6 \theta}}{5 - 4 \cos{2 \theta}}}$$
which, when expressed as an integral over the unit circle, we get, for the first integral
$$\frac{i}{2} \oint_{|z|=1} dz \frac{z}{(z^2-2)(2 z^2-1)}  $$
and for the second
$$\frac{i}{2} \oint_{|z|=1} dz \frac{z^7}{(z^2-2)(2 z^2-1)}  $$
Keep in mind that you will be taking the real part of the second integral when done.
