Find the value of $I$, using residue theorem. $$ I= \int_{\theta-\pi}^{\theta-2\pi} \frac{du}{r^2+1-2r\cos(u)} - \int_{\theta}^{\theta-\pi} \frac{du}{r^2+1-2r\cos(u)} $$ with $r<1$ and $0<\theta<2\pi$
Thanks.
I have used that $cos(u)=\frac{e^{iu}+e^{-iu}}{2}, z=e^{iu}, \frac{dz}{iz}=du$
Then, $$ I= \int_{C_1} \frac{dz}{z(r^2+1)-rz^2-r} - \int_{C_2} \frac{dz}{z(r^2+1)-rz^2-r} $$ where $C_1$ and $C_2$ are fixed with $\theta$.
I also have that the singular points of integrand are: $z_1=r, z_2=\frac{1}{r}$ but as $r<1$ I only consider $z_1$ to evaluate the Residue.
$$\operatorname{Res}\left(\frac{dz}{z(r^2+1)-rz^2-r}, z=r\right)=\lim_{z\to r} \frac{1}{r^2+1-2rz}=\frac{1}{1-r^2}$$
a) If $\theta=0, \pi, 2\pi,\ldots$ $\operatorname{Res}\left(\int_{C_1}\right)= \operatorname{Res}\left(\int_{C_2}\right)$ so $I=0$.
b) If $0<\theta<\pi$ singularity is only in $C_1$ then $I=\frac{1}{1-r^2}$
c) If $\pi<\theta<2\pi$ singularity is only in $C_2$ and $I=\frac{-1}{1-r^2}$
I am right? Please.