Q32 / 75 · JEE Main 2025 (3 Apr Shift 2) · Mathematics
The integral $\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x}$ is equal to
- Correct answerA$2 \pi^2$
- B$4 \pi^2$
- C$\pi^2$
- D$\frac{3 \pi^2}{2}$
Solution
$\begin{aligned}<br/>& I=\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x} \\
& I=\int_0^\pi \frac{8(\pi-x) d x}{4 \cos ^2 x+\sin ^2 x}<br/>\end{aligned}$
$2 \mathrm{I}=8 \pi \int_0^\pi \frac{d x}{4 \cos ^2 x+\sin ^2 x}$
$2 \mathrm{I}=8 \pi \times 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{4+\tan ^2 \mathrm{x}} \mathrm{dx}$
$\begin{aligned}<br/>& \mathrm{I}=8 \pi \int_0^{\infty} \frac{\mathrm{dt}}{4+\mathrm{t}^2}=8 \pi \times \frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{t}}{2}\right]_0^{\infty} \\ & =4 \pi \times \frac{\pi}{2}=2 \pi^2<br/>\end{aligned}$
option (1)
$2 \mathrm{I}=8 \pi \int_0^\pi \frac{d x}{4 \cos ^2 x+\sin ^2 x}$
$2 \mathrm{I}=8 \pi \times 2 \int_0^{\pi / 2} \frac{\sec ^2 x}{4+\tan ^2 \mathrm{x}} \mathrm{dx}$
$\begin{aligned}<br/>& \mathrm{I}=8 \pi \int_0^{\infty} \frac{\mathrm{dt}}{4+\mathrm{t}^2}=8 \pi \times \frac{1}{2}\left(\tan ^{-1} \frac{\mathrm{t}}{2}\right]_0^{\infty} \\ & =4 \pi \times \frac{\pi}{2}=2 \pi^2<br/>\end{aligned}$
option (1)
Hence, the correct answer is option: 'A'.