The value of the integral ${\int }_0^{\mathrm{\infty }} \frac{1}{a^2-2a\cos x+1}dx$ $(a<1)\text{ is }$

$I={\int }_0^{\mathrm{\infty }} \frac{1}{a^2-2a\cos x+1}dx$

Put $\cos x={\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}$

Divide above and below by ${\cos }^2\frac{x}{2}$ and put $\tan \frac{x}{2}=t$ and adjust the limit

$\therefore I=\frac{2}{(1+a{)}^2}{\int }_0^{\mathrm{\infty }} \frac{dt}{{\left(\frac{1-a}{1+a}\right)}^2+t^2}$

$\begin{array}{r}=\frac{2}{1-a^2}{\left[{\tan }^{-1}t\left(\frac{1+a}{1-a}\right)\right]}_0^{\mathrm{\infty }}\\ =\frac{2}{1-a^2}+\frac{\pi }{2}=\frac{\pi }{1-a^2}\end{array}$