$\int_0^{2\pi}\frac{d\theta}{a+b\cos\theta}$

The usual substitution: \(z=e^{i\theta}, \dtheta = \inverseof{(iz)} \dz\). \[ \int_{[0, 2\pi]} \inverseof{(a +b\cos(\theta))} \dtheta &= \oint \inverseof{\qty{ a + {b\over 2}(z+\inverseof{z})}} \inverseof{(iz)} \dz \\ &= -i\oint \inverseof{\qty{ za + {b\over 2}(z^2 + 1)}} \dz \\ &= -i \oint \inverseof{\qty{{b\over 2}z^2 + az + {b\over 2} }} \dz \\ &= -{2i\over b} \oint \inverseof{\qty{z^2 + {2a\over b}z + 1}} \dz \\ &= -{2i\over b}\oint \inverseof{(z-r_1)} \inverseof{(z-r_2)} \dz ,\] where by the quadratic formula the roots are \[ z_k &= {1\over 2} \qty{-{2a\over b} \pm \sqrt{\qty{2a\over b}^2 - 4}} \\ &= -{a\over b}\pm {1\over 2}\sqrt{{4a^2 \over b^2} - 4} \\ &= -{a\over b}\pm \sqrt{{a^2 \over b^2} - 1} \\ &= -{a\over b}\pm \sqrt{{a^2 - b^2 \over b^2}} \\ &= -{a\over b}\pm {1\over b } \sqrt{{a^2 - b^2}} .\] Thus \[ r_1 &\da \inverseof{b}\qty{-a + \sqrt{a^2-b^2}} \\ r_2 &\da \inverseof{b}\qty{-a - \sqrt{a^2-b^2}} .\]

Assume \(a>\abs b>0\), so that the integrand is continuous and the roots are real and distinct (for \(b=0\) the integral is \(2\pi/a\) directly). Since \(r_1 r_2 = 1\), exactly one root lies in \(\DD\), giving one simple pole there. If \(b>0\), then \(r_2=-(a+\sqrt{a^2-b^2})/b<-a/b<-1\); if \(b<0\), then \(r_2>a/\abs b>1\). In either case \(\abs{r_2}>1\), so \(r_1\in \DD\). Computing the residue here: \[ \Res_{z=r_1} \inverseof{(z-r_1)} \inverseof{(z-r_2)} &= \inverseof{(z-r_2)} \evalfrom_{z=r_1} \\ &= \inverseof{(r_1 - r_2)} \\ &= \inverseof{\qty{2\inverseof{b} \sqrt{a^2-b^2} }} ,\] so \[ I &= 2\pi i \cdot -{2 i \over b}{b\over 2\sqrt{a^2-b^2}} \\ &= {2\pi \over \sqrt{a^2-b^2} } .\]