Question:
The value of the integral $ \int^{ \pi / 2 }_ 0 \frac {\sqrt { \cot \, x }}{ \sqrt { \cot \, x } + \sqrt { \tan \, x }} \, dx \, is $
The value of the integral $ \int^{ \pi / 2 }_ 0 \frac {\sqrt { \cot \, x }}{ \sqrt { \cot \, x } + \sqrt { \tan \, x }} \, dx \, is $
Updated On: Jun 14, 2022
- $ \frac{\pi}{ 4 } $
- $ \frac{\pi}{ 2 } $
- $ \pi $
- None of these
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The Correct Option is A
Solution and Explanation
$ Let \, \, I = \int^{ \pi / 2 }_ 0 \frac {\sqrt { cot \, x }}{ \sqrt { cot \, x } + \sqrt { tan \, x }} \, dx \, is ..................(1)$
$ \Rightarrow \, \, I = \int^{ \pi / 2 }_ 0 \frac {\sqrt { tan \, x }}{ \sqrt { cot \, x } + \sqrt { tan \, x }} \, dx \, is ..................(2)$
On adding Eqs. (i) and (ii), we get
$ 2I = \int^{ \pi / 2 }_ 0 \, 1 \, dx $
$ \therefore I = \frac{\pi}{4 } $
$ \Rightarrow \, \, I = \int^{ \pi / 2 }_ 0 \frac {\sqrt { tan \, x }}{ \sqrt { cot \, x } + \sqrt { tan \, x }} \, dx \, is ..................(2)$
On adding Eqs. (i) and (ii), we get
$ 2I = \int^{ \pi / 2 }_ 0 \, 1 \, dx $
$ \therefore I = \frac{\pi}{4 } $
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