STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

$ = \int_0^\pi {\frac{{dx}}{{{{(1 - a)}^2}{{\cos }^2}\frac{x}{2} + {{(1 + a)}^2}{{\sin }^2}\frac{x}{2}}}} $

$ = \frac{2}{{{{(1 + a)}^2}}}\int_0^\infty {\frac{{dt}}{{{{\left\{ {(1 - a)/(1 + a)} \right\}}^2} + {t^2}}}} $;                               {जहाँ $t = \tan \frac{x}{2}$}

$ = \frac{2}{{{{(1 + a)}^2}}}\frac{{(1 + a)}}{{(1 - a)}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{1 + a}}{{1 - a}}.t} \right)} \right]_0^\infty $

$ = \frac{2}{{(1 - {a^2})}}[{\tan ^{ - 1}}\infty - {\tan ^{ - 1}}0] = \frac{\pi }{{1 - {a^2}}}$.