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The value of the integral \(\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \frac{{\sqrt {tanx} }}{{1 + \sqrt {tanx} }}\) dx is
1. \(\frac{\pi}{2}\)
2. \(\frac{\pi}{4}\)
3. \(\frac{\pi}{6}\)
4. None of these
5. \(\frac{\pi}{3}\)

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Best answer
Correct Answer - Option 2 : \(\frac{\pi}{4}\)

CONCEPT:

\(\mathop \smallint \nolimits_a^b f\left( x \right)\;dx = \;\mathop \smallint \nolimits_a^b f\left( {a + b - x} \right)\;dx\)

CALCULATION:

Here, we have to find the value of the integral \(\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \frac{{\sqrt {tanx} }}{{1 + \sqrt {tanx} }}\)

Let \(I = \mathop \smallint \nolimits_0^{\frac{\pi }{2}} \frac{{\sqrt {tanx} }}{{1 + \sqrt {tanx} }}dx\)       ------(1)

As we know that, \(\mathop \smallint \nolimits_a^b f\left( x \right)\;dx = \;\mathop \smallint \nolimits_a^b f\left( {a + b - x} \right)\;dx\)

⇒ \(I = \mathop \smallint \nolimits_0^{\frac{\pi }{2}} \frac{{\sqrt {cotx} }}{{1 + \sqrt {cotx} }} dx\)       ------(2)

Adding equation (1) and (2), we get

⇒ \(2I = \;\mathop \smallint \nolimits_0^{\frac{\pi }{2}\;} \left[ {\frac{{\sqrt {\tan x} }}{{1 + \sqrt {\tan x} }} + \frac{{\sqrt {\cot x} }}{{1 + \sqrt {\cot x} }}} \right] dx\)

⇒ \(2I = \;\mathop \smallint \limits_0^{\frac{\pi }{2}} dx = \frac{\pi }{2}\;\)

⇒ \(I = \frac{\pi}{4}\)

Hence, option B is the correct answer.

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