<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.
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<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>
Correct Answer: A