<p>Evaluate \(\displaystyle\int_0^{\pi/2}\frac{dx}{1+\sqrt{\tan x}}\) [JEE Main 2022]</p>
Step-by-Step Solution
Key Concept: King: I = \int\sqrt{tanx}/(1+\sqrt{tanx})dx. Add: 2I = \int_0^(\pi/2)1 dx = \pi/2 \to I = \pi/4.
<div class='solution'>
<p>Let \(I=\int_0^{\pi/2}\frac{dx}{1+\sqrt{\tan x}}\). King (\(x\to\pi/2-x\)): \(\tan(\pi/2-x)=\cot x=1/\tan x\).</p>
<p>\[I=\int_0^{\pi/2}\frac{dx}{1+\sqrt{\cot x}}=\int_0^{\pi/2}\frac{\sqrt{\tan x}}{{\sqrt{\tan x}+1}}dx\]</p>
<p>Add: \(2I=\int_0^{\pi/2}\frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}}dx=\frac{\pi}{2}\Rightarrow I=\boxed{\frac{\pi}{4}}\)</p>
</div>
Correct Answer: A