x ;dx 2x = - Vidyadip

MCQ

$\int {\frac{{\sec x\;dx}}{{\sqrt {\cos 2x} }}} = $

✓
${\sin ^{ - 1}}(\tan x)$
B
$\tan x$
C
${\cos ^{ - 1}}(\tan x)$
D
$\frac{{\sin x}}{{\sqrt {\cos x} }}$

✓

Answer

Correct option: A.

${\sin ^{ - 1}}(\tan x)$

a
(a)$\int_{}^{} {\frac{{\sec x\,dx}}{{\sqrt {\cos 2x} }}} = \int_{}^{} {\frac{{\sec x}}{{\sqrt {{{\cos }^2}x - {{\sin }^2}x} }}} \,dx$
$ = \int_{}^{} {\frac{{{{\sec }^2}x\,dx}}{{\sqrt {1 - {{\tan }^2}x} }}} $ $\{$Multiplying $N'r$ and $D'r$ by $\sec x\} $
Now putting $\tan x = t \Rightarrow {\sec ^2}x\,dx = dt,$

we get the integral $ = {\sin ^{ - 1}}t = {\sin ^{ - 1}}(\tan x).$
Trick : Since $\frac{d}{{dx}}\{ {\sin ^{ - 1}}(\tan x)\} = \frac{{{{\sec }^2}x}}{{\sqrt {1 - {{\tan }^2}x} }}$
$ = \frac{{{{\sec }^2}x.\cos x}}{{\sqrt {{{\cos }^2}x - {{\sin }^2}x} }} = \frac{{\sec x}}{{\sqrt {\cos 2x} }}.$

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