# The value of the integral $\mathop \smallint \nolimits_0^{\frac{\pi }{2}} \frac{{\sqrt {tanx} }}{{1 + \sqrt {tanx} }}$ dx is

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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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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.