_ ^ x x x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{\sqrt {\tan x} }}{{\sin x\cos x}}} \;dx = $

A
$2\sqrt {\sec x} + c$
✓
$2\sqrt {\tan x} + c$
C
$\frac{2}{{\sqrt {\tan x} }} + c$
D
$\frac{2}{{\sqrt {\sec x} }} + c$

✓

Answer

Correct option: B.

$2\sqrt {\tan x} + c$

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

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