<p>\(\displaystyle\int_0^{\pi/4}\frac{\sec^2 x}{(1+\tan x)^2}\,dx\) [JEE Main 2019]</p>
Step-by-Step Solution
Key Concept: Let t = tan x, dt = sec^2x dx. Limits 0\to 1. \int_0^1 dt/(1+t)^2 = [-1/(1+t)]_0^1 = -1/2+1 = 1/2.
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<p>Let $t=\tan x\Rightarrow dt=\sec^2 x\,dx$. Limits: $x=0\to t=0$; $x=\pi/4\to t=1$.</p>
<p>$$I=\int_0^1\frac{dt}{(1+t)^2}=\left[\frac{-1}{1+t}\right]_0^1=-\frac{1}{2}+1=\boxed{\frac{1}{2}}$$</p>
Correct Answer: A