MCQ
$\int_0^\pi {\frac{{dx}}{{1 - 2a\cos x + {a^2}}}} \, =$
- A$\frac{\pi }{{2(1 - {a^2})}}$
- B$\pi (1 - {a^2})$
- ✓$\frac{\pi }{{1 - {a^2}}}$
- DNone of these
$ = \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}}}} $; {where $t = \tan \frac{x}{2}$}
$ = \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}}}$.
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$(I)$ $g$ is an increasing function in $(0,1)$
$(II)$ $g$ is one-one in $(0,1)$ Then,