Definite Integration Question 101
Question: $ \int_0^{\pi /2}{\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}dx=} $
[MP PET 1990, 95; IIT 1983; MNR 1990]
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{4} $
D) $ \frac{\pi }{3} $
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Answer:
Correct Answer: C
Solution:
$ I=\int_0^{\pi /2}{\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}dx} $ …..(i)
$ =\int_0^{\pi /2}{\frac{\sqrt{\cot ( \frac{\pi }{2}-x )}}{\sqrt{\cot ( \frac{\pi }{2}-x )}+\sqrt{\tan ( \frac{\pi }{2}-x )}}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}} $ .
BETA
