Question

${\int }_0^{\pi }\frac{dx}{1-2acosx+a^2}$, $a>1$

Answer 1

$\begin{array}{c}{\int }_0^{\pi }\frac{dx}{\left(1-2a\cos x+a^2\right)}a>1\\ ={\int }_0^{\pi }\frac{dx}{1-2a\frac{\left(1-{\tan }^2\frac{x}{2}\right)}{\left(1+{\tan }^2\frac{x}{2}\right)}+a^2}\\ ={\int }_0^{\pi }\frac{{\sec }^2\frac{x}{2}dx}{\left(a^2+1\right)\left(1+{\tan }^2\frac{x}{2}\right)-2a\left(1-{\tan }^2\frac{x}{2}\right)}\\ put\tan \frac{x}{2}=t\\ \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt\\ ={\int }_0^{\infty }\frac{2dt}{\left(a^2+1\right)\left(1+t^2\right)-2a\left(1-t^2\right)}\\ ={\int }_0^{\infty }\frac{dt}{\left(1-a^2\right)+\left(1+a^2\right)t^2}\\ =\frac{2}{1-a}\times \frac{\pi }{2}\\ =\frac{\pi }{1-a}\\ whichistherequiredanswer.\end{array}$