JEE Main 2025 — Definite Integration Question with Solution
JEE Main 2025 (3 Apr Shift 2)
Question
The integral is equal to
Choose an option
Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
$\begin{aligned}
& I=\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x} \\ & I=\int_0^\pi \frac{8(\pi-x) d x}{4 \cos ^2 x+\sin ^2 x}
\end{aligned}2 \mathrm{I}=8 \pi \int_0^\pi \frac{d x}{4 \cos ^2 x+\sin ^2 x}2 \mathrm{I}=8 \pi \times 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{4+\tan ^2 \mathrm{x}} \mathrm{dx}\begin{aligned}
& \mathrm{I}=8 \pi \int_0^{\infty} \frac{\mathrm{dt}}{4+\mathrm{t}^2}=8 \pi \times \frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{t}}{2}\right]_0^{\infty} \\ & =4 \pi \times \frac{\pi}{2}=2 \pi^2
\end{aligned}$
option (1)
& I=\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x} \\ & I=\int_0^\pi \frac{8(\pi-x) d x}{4 \cos ^2 x+\sin ^2 x}
\end{aligned}2 \mathrm{I}=8 \pi \int_0^\pi \frac{d x}{4 \cos ^2 x+\sin ^2 x}2 \mathrm{I}=8 \pi \times 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{4+\tan ^2 \mathrm{x}} \mathrm{dx}\begin{aligned}
& \mathrm{I}=8 \pi \int_0^{\infty} \frac{\mathrm{dt}}{4+\mathrm{t}^2}=8 \pi \times \frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{t}}{2}\right]_0^{\infty} \\ & =4 \pi \times \frac{\pi}{2}=2 \pi^2
\end{aligned}$
option (1)
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This is a previous-year question from JEE Main 2025, covering the Definite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.