Question

Solve :

${\int }_0^{\pi }\frac{1}{a^2-2a\cos x+1}dx$

Answer 1

$I={\int }_0^{\pi }\frac{1}{a^2-2a\cos x+1}dxputtingthevalueof\cos xas\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\therefore I={\int }_0^{\pi }\frac{1}{a^2-2a.\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+1}dx={\int }_0^{\pi }\frac{1+{\tan }^2\frac{x}{2}}{(a^2+1)(1+{\tan }^2\frac{x}{2})-2a\left(1{-\tan }^2\frac{x}{2}\right)}dx={\int }_0^{\pi }\frac{{\sec }^2\frac{x}{2}}{(a^2+1)(1+{\tan }^2\frac{x}{2})-2a\left(1{-\tan }^2\frac{x}{2}\right)}dx$

$Let\tan \frac{x}{2}=t\Rightarrow {\sec }^2\frac{x}{2}dx=2dtalso,x=0\Rightarrow t=\tan 0=0and,x=\pi \Rightarrow t=\tan \frac{\pi }{2}=\infty \therefore I={\int }_0^{\infty }\frac{2dt}{(a^2+1)(1+t^2)-2a(1-t^2)}=2{\int }_0^{\infty }\frac{dt}{{(1-a)}^2+{\left(\right(1+a\left)t\right)}^2}=\frac{2}{{(1-a)}^2}{\int }_0^{\infty }\frac{dt}{1+{\left[\frac{1+a}{1-a}t\right]}^2}againLet\frac{1+a}{1-a}t=u\Rightarrow dt=du\left(\frac{1-a}{1+a}\right)\therefore I=\frac{2}{(1-a)(1+a)}{\int }_0^{\infty }\frac{du}{1+u^2}=\frac{2}{(1-a)(1+a)}{\left[{\tan }^{-1}u\right]}_0^{\infty }=\frac{2}{1-a^2}(\frac{\pi }{2}-0)=\frac{\pi }{1-a^2}$