$\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$
Let z=$e^{i\theta}$$\Rightarrow$$\cos\theta=\frac{z^2+1}{2z}$ $\Rightarrow$ $a+b\cos\theta=a+\frac{b}{2}(z+\frac{1}{z})$,$d\theta=\frac{dz}{iz}$
$\int_{0}^{2\pi}\frac{d\theta}{a+b\cos\theta}$
=$\int_{|z|=1}\frac{dz}{iz(a+\frac{b}{2}(z+\frac{1}{z}))}$
=$\int_{|z|=1}\frac{2dz}{i(2az+bz^2+b)}$
$2az+bz^2+b=0$$\Rightarrow$ $z=\frac{-a\pm\sqrt{a^{2}-b^{2}}}{b}$
a>b>0$\Rightarrow$$\frac{-a}{b}<-1$ $\Rightarrow \frac{-a+ \sqrt{a^{2}-b^{2}}}{b} >-1 $which is inside |z|=1 and $\frac{-a- \sqrt{a^{2}-b^{2}}}{b}$<-1 which is outside |z|=1
$\therefore$ $\int_{|z|=1}\frac{2dz}{i(2az+bz^2+b)}$
=$\int_{|z|=1}\frac{2/i(z+\frac{a+\sqrt{a^{2}-b^{2}}}{b})}{(z-\frac{-a+\sqrt{{a^2}-{b^2}}}{b})}dz$
=$2\pi i f(z_0)$
=$2\pi i \frac{2}{\frac{2i\sqrt{{a^2}-{b^2}}}{b}}$
=$\frac{2\pi b}{\sqrt{{a^2}-{b^2}}} $
