Answer is: 12
\(=24 \int_{0}^{\frac{\pi}{48}}-\sin \left(4 \mathrm{x}-\frac{\pi}{12}\right)+\int_{\pi / 48}^{\pi / 4} \sin \left(4 \mathrm{x}-\frac{\pi}{12}\right)\)
\(+\int_{0}^{\frac{\pi}{6}}[0] \mathrm{dx}+\int_{\pi / 6}^{\pi / 4}\left[2 \sin _{\downarrow} \mathrm{x}\right] \mathrm{dx} \)
\(= 24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right]+\frac{\pi}{4}-\frac{\pi}{6}\)
\(=24\left(\frac{1}{2}+\frac{\pi}{12}\right)=2 \pi+12\)
\(\alpha=12\)