_ ^ 1 1 + ^2 x dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{1 + {{\cos }^2}x}}dx} = $

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

✓

Answer

Correct option: C.

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

c
(c)$\int_{}^{} {\frac{{dx}}{{1 + {{\cos }^2}x}}} = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\sec }^2}x + 1}}} = \int_{}^{} {\frac{{{{\sec }^2}x}}{{{{\tan }^2}x + 2}}} \,dx$
$ = \int_{}^{} {\frac{{dt}}{{{t^2} + 2}} = \frac{1}{{\sqrt 2 }}{{\tan }^{ - 1}}\left( {\frac{t}{{\sqrt 2 }}} \right) + c} $ $\{$Putting $\tan x = t\} $
$ = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}\tan x} \right) + c$.
Trick : By inspection,
$\frac{d}{{dx}}\left\{ {\frac{1}{{\sqrt 2 }}{{\tan }^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}\tan x} \right)} \right\} = \frac{1}{{\sqrt 2 }}\left( {\frac{1}{{1 + \frac{{{{\tan }^2}x}}{2}}}} \right)\frac{1}{{\sqrt 2 }}{\sec ^2}x$
$ = \frac{1}{2}\,.\,\frac{{2{{\sec }^2}x}}{{(2 + {{\tan }^2}x)}} = \frac{{{{\sec }^2}x}}{{1 + {{\sec }^2}x}} = \frac{1}{{1 + {{\cos }^2}x}}$.

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