I'm not sure how to solve this integral: $$ \int_{0}^{2\pi} [\frac{a+\cos(n\theta)}{a^2+1+2acos(n\theta)}] \ d\theta $$
SUGGGESTION: Use the function $ f(z)=\frac{1}{z^n+a} $
Solution (I developed some steps but I dont know how can I continue)
Let $C(0,1)$ be the unit circle $\mathbb{C}$, the integral: $$ \int_{C(0,1)}f(z)dz$$ Using the standard parametrization of $C(0,1)$: $$ \int_{C(0,1)}f(z)dz = \int_{0}^{2\pi}f(e^{ix})(ie^{ix})dx $$ Note that $$cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$ The integral becomes
$$ \int_{0}^{2\pi} [\frac{a+\cos(n\theta)}{a^2+1+2acos(n\theta)}] \ d\theta = \int_{0}^{2\pi} \frac{a+\frac{e^{in\theta}-e^{-in\theta}}{2}}{a^2+1+2a\frac{e^{ix}-e^{-ix}}{2}} $$ Now lets use the substitution $z=e^{i\theta}$, then $dz=ie^{i\theta}d\theta=izd\theta$ hence $d\theta=\frac{dz}{iz}$:
$$\begin{align} \int_{0}^{2\pi}\frac{a+\frac{e^{in\theta}+e^{-in\theta}}{2}}{a^2+1+2a\frac{e^{ix}+e^{-ix}}{2}} =\int_{C(0,1)}\frac{a+\frac{z^n+z^{-n}}{2}}{a^2+1+2a\frac{z^n+z^{-n}}{2}} =\int_{C(0,1)}\frac{az^n+z^{(2n)}+1}{a^2z^n+z^n+az^{2n}+a} \\ =\int_{C(0,1)}\frac{az^n+z^{(2n)}+1}{(a+z^n)(az^n+1)} \end{align}$$
From here, I dont know how I can continue
Suppose we seek to evaluate $$\int_0^{2\pi} \frac{a+\cos(n\theta)}{a^2+1+2a\cos(n\theta)} d\theta.$$ where $a\gt 1.$
Introduce $z=\exp(ni\theta)$ so that $dz=niz \; d\theta$ to get (observe that we trace the unit circle exactly $n$ times)
$$n\int_{|z|=1} \frac{a+(z+1/z)/2}{a^2+1+2a(z+1/z)/2} \frac{dz}{niz} \\ = \frac{1}{i} \int_{|z|=1} \frac{2az+z^2+1}{2a^2z+2z+2az^2+2a} \frac{dz}{z}.$$
Now put
$$f(z) = \frac{z^2+2az+1}{2az^2+2(a^2+1)z+2a} \frac{1}{z}.$$
We have $$\mathrm{Res}_{z=0} f(z) = \frac{1}{2a}.$$
The two roots of the denominator are at $z=-a$ and $z=-1/a$ and with $a\gt 1$ only the second one is inside the contour. We get
$$\mathrm{Res}_{z=-1/a} f(z) = -a\frac{1/a^2-2+1}{4a(-1/a)+2(a^2+1)} = -a \frac{1/a^2-1}{2a^2-2} \\ = -\frac{1}{2a} \frac{1-a^2}{a^2-1} = \frac{1}{2a}.$$
This finally yields for the value of the integral
$$\frac{1}{i}\times 2\pi i\times \frac{1}{a} = \frac{2\pi}{a}.$$
For the other case $0\lt a\lt 1$ we get the residue
$$\mathrm{Res}_{z=-a} f(z) = -\frac{1}{a}\frac{a^2-2a^2+1}{-4a^2+2(a^2+1)} = -\frac{1}{a}\frac{1-a^2}{-2a^2+2} \\ = \frac{1}{2a}\frac{1-a^2}{a^2-1} = -\frac{1}{2a}.$$
We see that the integral is zero in this case.
The cases $a=0$ and $a=1$ can be done with basic algebra and are left to the reader.