Evaluate the integral $\int_{0}^{2\pi} \frac {\cos^2(x)}{13+12\cos(x)} \,dx$ using the residue theorem.
I have managed to make a start on this problem by putting this problem in a complex analysis setting i.e. by changing the integral to $\int_{\gamma} -\frac {\frac {i}{4} (z+\frac {1}{z})^2}{\left(13+6z+\frac {6}{z}\right)z} \,dz$ where $\gamma$ is the path $ t \mapsto e^{it}$. I then found that the singularities of the function $f(z) = -\frac {\frac {i}{4} \left(z+\frac {1}{z}\right)^2}{\left(13+6z+\frac {6}{z}\right)z}$ are at $z=0,z=-\frac {2}{3}, z = -\frac {3}{2}$.
I know that I need to now calculate the residues for $z=0, z= -\frac {2}{3}$ (and not $z=-\frac {3}{2}$ since this singularity lies 'outside' the path) however I am unsure of how to do this.
You want to compute$$\int_0^{2\pi}R(\cos\theta)\,\mathrm d\theta,\tag1$$with $\displaystyle R(x)=\frac{x^2}{13+12x}$. So, consider\begin{align}\frac1zR\left(\frac{z+1/z}2\right)&=\frac{(z^2+1)^2}{4z^2(6z^2+13z+6)}\\&=\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)},\end{align}and then $(1)$ is equal to$$2\pi\left(\operatorname{res}_{z=0}\left(\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)}\right)+\operatorname{res}_{z=-2/3}\left(\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)}\right)\right).$$In order to compute$$\operatorname{res}_{z=0}\left(\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)}\right),\tag2$$define$$g(z)=\frac{(z^2+1)^2}{24\left(z+\frac23\right)\left(z+\frac32\right)},$$and then $(2)=g'(0)=-\frac{13}{144}$. And, in order to compute$$\operatorname{res}_{z=-2/3}\left(\frac{(z^2+1)^2}{24z^2\left(z+\frac23\right)\left(z+\frac32\right)}\right),\tag3$$define$$h(z)=\frac{(z^2+1)^2}{24z^2\left(z+\frac32\right)}=\frac{(z^2+1)^2}{12z^2(2z+3)},$$and then $(3)=h\left(-\frac23\right)=-\frac{169}{720}$. So, $(1)$ is equal to $2\pi\left(-\frac{13}{144}+\frac{169}{720}\right)=2\pi\frac{13}{90}=\pi\frac{13}{45}$.