_ ^ dx 2 + x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{2 + \cos x}} = } $

A
$2{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
✓
$\frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
C
$\frac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 3 }}\tan \frac{x}{2}} \right) + c$
D
None of these

✓

Answer

Correct option: B.

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

b
(b)$\int_{}^{} {\frac{{dx}}{{2 + \cos x}} = \int_{}^{} {\frac{{dx}}{{2{{\sin }^2}\left( {\frac{x}{2}} \right) + 2{{\cos }^2}\left( {\frac{x}{2}} \right) + {{\cos }^2}\left( {\frac{x}{2}} \right) - {{\sin }^2}\left( {\frac{x}{2}} \right)}}} } $
$ = \int_{}^{} {\frac{{dx}}{{{{\sin }^2}\left( {\frac{x}{2}} \right) + 3{{\cos }^2}\left( {\frac{x}{2}} \right)}}} = \int_{}^{} {\frac{{{{\sec }^2}\left( {\frac{x}{2}} \right)}}{{{{\tan }^2}\left( {\frac{x}{2}} \right) + 3}}dx} $
Put $\tan \left( {\frac{x}{2}} \right) = t \Rightarrow {\sec ^2}\left( {\frac{x}{2}} \right)\,dx = 2dt,$ then it reduces to
$2\int_{}^{} {\frac{{dt}}{{{t^2} + 3}}} = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{t}{{\sqrt 3 }}} \right) + c = \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{\tan \left( {\frac{x}{2}} \right)}}{{\sqrt 3 }}} \right) + c$.

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