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

MCQ

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

✓
$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
B
$\sqrt 2 {\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
C
$ - \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$
D
$ - \sqrt 2 {\tan ^{ - 1}}(\sqrt 2 \tan x) + k$

✓

Answer

Correct option: A.

$\frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$

a
(a) $I = \int_{}^{} {\frac{1}{{1 + {{\sin }^2}x}}\,dx} = \int_{}^{} {\frac{{dx}}{{2{{\sin }^2}x + {{\cos }^2}x}}} $
$ = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{2{{\tan }^2}x + 1}}} $$ = \frac{1}{2}\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\tan }^2}x + \frac{1}{2}}}} $
Put $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt,$ then
$I = \frac{1}{2}\int_{}^{} {\frac{{dt}}{{{t^2} + \frac{1}{2}}} = \frac{1}{2}} .\frac{1}{{1\sqrt 2 }}{\tan ^{ - 1}}\frac{t}{{1\sqrt 2 }}$
$ = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}(\sqrt 2 \tan x) + k$.

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