Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$ 
I want to prove that $\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$

My ideas, I don't know if they lead anywhere:
Let's substitute $\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and $z=e^{i\theta}$ right after:
$\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=-i\cdot\int_1^{-1}\frac{z^2+z^{-2}+\frac{1}{2}z^3+\frac{1}{2}z^{-3}}{5z+2z^2+2}dz$
This now gives me 4 new integrals, for example 
$\displaystyle-i\int_1^{-1}\frac{z^2}{2z^2+5z+2}dz$, $\displaystyle-i\int_1^{-1}\frac{1}{2z^4+5z^3+2z^2}dz$ and so on.
But since I haven't been able to solve any of the new integrals, I'm a little lost. 

Edit: Can't I do a partial fractions decomposition of all the 4 integrals and solve them seperately?


If you wish to exploit the residue theorem, then first exploit the fact that the integral is even.  In addition, use Euler's formula to write $2\cos(2\theta)+\cos(3\theta)=\text{Re}(2e^{i2\theta}+e^{i3\theta})$.
Then, we have
$$\begin{align}
\int_0^\pi \frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}\,d\theta&=\frac12\text{Re}\left(\oint_{|z|=1}\frac{2z^2+z^3}{5+2(z+z^{-1})}\,\frac{1}{iz}\,dz\right)\\\\
&=\frac12\text{Re}\left(\frac1i\oint_{|z|=1}\frac{z^2(2+z)}{(2z+1)(z+2)}\,dz\right)\\\\
&=\frac12\text{Re}\left(\frac1i\oint_{|z|=1}\frac{z^2}{2z+1}\,dz\right)\\\\
&=\frac{\pi}{8} 
\end{align}$$
And we are done!
