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.