_0^ /2 dx 2 + x = - Vidyadip

MCQ

$\int_0^{\pi /2} {\frac{{dx}}{{2 + \cos x}}} = $

A
$\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$
B
$\sqrt 3 {\tan ^{ - 1}}\left( {\sqrt 3 } \right)$
✓
$\frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$
D
$2\sqrt 3 {\tan ^{ - 1}}\left( {\sqrt 3 } \right)$

✓

Answer

Correct option: C.

$\frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)$

c
(c) $I = \int_0^{\pi /2} {\frac{{dx}}{{2 + \cos x}}} $

$ = \int_0^{\pi /2} {\frac{{dx}}{{2{{\sin }^2}\frac{x}{2} + 2{{\cos }^2}\frac{x}{2} + {{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}}} $

$ = \int_0^{\pi /2} {\frac{{dx}}{{{{\sin }^2}\frac{x}{2} + 3{{\cos }^2}\frac{x}{2}}}} $

$= \int_0^{\pi /2} {\frac{{{{\sec }^2}\frac{x}{2}}}{{3 + {{\tan }^2}\frac{x}{2}}}dx} $

Put $t = \tan \frac{x}{2} $

$\Rightarrow dt = \frac{1}{2}{\sec ^2}\frac{x}{2}dx$, then

$I = 2\int_0^1 {\frac{{dt}}{{3 + {t^2}}} = \frac{2}{{\sqrt 3 }}{{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}} \right)} $.

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