JEE Main 2025 — 29 January, Evening Shift — Question 62

$=24 \int_{0}^{\frac{\pi}{48}}-\sin \left(4 \mathrm{x}-\frac{\pi}{12}\right)+\int_{\pi / 48}^{\pi / 4} \sin \left(4 \mathrm{x}-\frac{\pi}{12}\right)$ $\begin{aligned} & +\int_{0}^{\frac{\pi}{6}}[0] \mathrm{dx}+\int_{\pi / 6}^{\pi / 4}\left[2 \sin \mathrm{x}\right] \mathrm{dx} \\ = & 24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right]+\frac{\pi}{4}-\frac{\pi}{6} \end{aligned}$

$=24\left(\frac{1}{2}+\frac{\pi}{12}\right)=2 \pi+12$

$\alpha=12$