MCQ
$\int{\frac{{\sec x\;dx}}{{\sqrt {\cos 2x} }}} = $
- ✓${\sin ^{ - 1}}(\tan x)$
- B$\tan x$
- C${\cos ^{ - 1}}(\tan x)$
- D$\frac{{\sin x}}{{\sqrt {\cos x} }}$
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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