How do you evaluate the following integrals?
$$\int_0^{2\pi}\frac{1}{R^2+2R\cos\theta+1} \mathrm d\theta$$
$$\int_0^{2\pi}\frac{R\cos\theta}{R^2+2R\cos\theta+1} \mathrm d\theta$$
$$\int_0^{2\pi}\frac{R\sin\theta}{R^2+2R\cos\theta+1} \mathrm d\theta$$
Apparently there are three cases: one case for $R>1$, another case for $R<1$, and another case for $R=1$.
For context: I am trying to answer problem 6.3.4 of Arfken & Weber (2005). Mathematical Methods for Physicists which is to show that
$$\oint_C \frac{\mathrm dz}{z^2+z}=0$$ in which the contour $C$ is defined as $|z|=R>1$ (without using Cauchy's integral formula or the residue theorem).
These integrals are related to the Poisson kernel. If $R>1$ we have $$ \frac{1}{R+e^{i\theta}} = \sum_{n\geq 0}\frac{(-1)^n e^{ni\theta}}{R^{n+1}},\qquad \frac{1}{R+e^{-i\theta}} = \sum_{n\geq 0}\frac{(-1)^n e^{-ni\theta}}{R^{n+1}} $$ hence by considering the sum/difference/product between these representations, and the fact that $\int_{0}^{2\pi} e^{ki\theta}\,d\theta = 2\pi \delta(k)$, the computation of the given integrals is straightforward. If $R\in(0,1)$ one may apply the same principle to $$ \frac{1}{R+e^{i\theta}} = \sum_{n\geq 0}(-1)^n R^n e^{-(n+1)i\theta},\qquad \frac{1}{R+e^{-i\theta}} = \sum_{n\geq 0}(-1)^n R^n e^{(n+1)i\theta}$$ and if $R=1$ one may tackle the given integrals through the tangent half-angle substitution.