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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