JEE Main 2025 — 3 April, Evening Shift — Question 24

$I=\int_{0}^{\pi} \frac{8 x}{4 \cos ^{2} x+\sin ^{2} x} d x …(1)$

$I=\int_{0}^{\pi} \frac{8(\pi-x)}{4 \cos ^{2}(\pi-x)+\sin ^{2}(\pi-x)} d x$

$$\begin{gathered} I=\int_{0}^{\pi} \frac{8(\pi-x)}{4 \cos ^{2} x+\sin ^{2} x} d x …(2) \end{gathered}$$

Adding (1) and (2)

$2 I=8 \pi \int_{0}^{\pi} \frac{1}{4 \cos ^{2} x+\sin ^{2} x} d x$

$I=4 \pi \times 2 \int_{0}^{\pi} \frac{\sec ^{2} x}{4 \tan ^{2} x} d x$

Put $\tan x=t$ $\sec ^{2} x d x=d t$

$I=8 \pi \int_{0}^{\infty} \frac{d t}{4+t^{2}}$

$$\begin{aligned} & I=8 \pi \frac{1}{2}\left(\tan ^{-1} \frac{t}{2}\right)_{0}^{\infty} \\& I=4 \pi\left(\frac{\pi}{2}\right) \\& I=2 \pi^{2} \end{aligned}$$