_ ^ dx 1 + 3 ^2 x = - Vidyadip

MCQ

$\int_{}^{} {\frac{{dx}}{{1 + 3{{\sin }^2}x}} = } $

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

✓

Answer

Correct option: B.

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

b
(b)$\int_{}^{} {\frac{{dx}}{{1 + 3{{\sin }^2}x}}} = \int_{}^{} {\frac{{dx}}{{{{\sin }^2}x + {{\cos }^2}x + 3{{\sin }^2}x}}} $
$ = \int_{}^{} {\frac{{dx}}{{4{{\sin }^2}x + {{\cos }^2}x}}} = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{4{{\tan }^2}x + 1}} = \frac{1}{4}\int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{{{\tan }^2}x + \frac{1}{4}}}} } $
Put $t = \tan x \Rightarrow dt = {\sec ^2}x\,dx,$ then it reduces to
$\frac{1}{4}\int_{}^{} {\frac{{dt}}{{{t^2} + {{\left( {\frac{1}{2}} \right)}^2}}}} = \frac{1}{4}2{\tan ^{ - 1}}(2t) + c$
$ = \frac{1}{2}{\tan ^{ - 1}}(2t) + c = \frac{1}{2}{\tan ^{ - 1}}(2\tan x) + c.$

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