<s>[INST] 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?

 [/INST] As Dr. Sonnhard Graubner answered, expand $\cos(2x)$ and $\cos(3x)$ using the classical formulas and use the tangent half-angle substitution to get 
$$I=\int\frac{\cos(2x)+\cos(3x)}{5+4\cos(x}\,dx=\int\frac{2 \left(t^6+5 t^4-25 t^2+3\right)}{\left(t^2+1\right)^3 \left(t^2+9\right)}dt$$ Using partial fraction decomposition $$\frac{2 \left(t^6+5 t^4-25 t^2+3\right)}{\left(t^2+1\right)^3 \left(t^2+9\right)}=\frac{3}{8 \left(t^2+9\right)}+\frac{13}{8
   \left(t^2+1\right)}-\frac{9}{\left(t^2+1\right)^2}+\frac{8}{\left(t^2+1\right)^3
   }$$ Integration now to get $$I=\frac{1}{8} \left(\frac{4 t-12 t^3}{\left(t^2+1\right)^2}+\tan
   ^{-1}\left(\frac{t}{3}\right)+\tan ^{-1}(t)\right)$$ and the bounds are $0$ and $\infty$.
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