_0^ /2 x x + x ,dx = - Vidyadip

MCQ

$\int_0^{\pi /2} {\frac{{\sqrt {\cot x} }}{{\sqrt {\cot x} + \sqrt {\tan x} }}\,dx = } $

A
$\pi $
B
$\frac{\pi }{2}$
✓
$\frac{\pi }{4}$
D
$\frac{\pi }{3}$

✓

Answer

Correct option: C.

$\frac{\pi }{4}$

c
(c) $I = \int_0^{\pi /2} {\frac{{\sqrt {\cot x} }}{{\sqrt {\cot x} + \sqrt {\tan x} }}dx} $.....$(i)$

$ = \int_0^{\pi /2} {\frac{{\sqrt {\cot \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\cot \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\tan \left( {\frac{\pi }{2} - x} \right)} }}dx} $

$ = \int_0^{\pi /2} {\frac{{\sqrt {\tan x} }}{{\sqrt {\tan x} + \sqrt {\cot x} }}dx} $.....$(ii)$

Now adding $(i)$ and $(ii),$ we get

$2I = \int_0^{\pi /2} {\frac{{\sqrt {\cot x} + \sqrt {\tan x} }}{{\sqrt {\tan x} + \sqrt {\cot x} }}dx = [x]_0^{\pi /2} \Rightarrow I = \frac{\pi }{4}} $.

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