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If \(24 \int_{0}^{\frac{\pi}{4}}\left(\sin \left|4 x-\frac{\pi}{12}\right|+[2 \sin x]\right) d x=2 \pi+\alpha,\) where [•] denotes the greatest integer function, then \(\alpha\) is equal to _______.

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Answer is: 12 

\(=24 \int_{0}^{\frac{\pi}{48}}-\sin \left(4 \mathrm{x}-\frac{\pi}{12}\right)+\int_{\pi / 48}^{\pi / 4} \sin \left(4 \mathrm{x}-\frac{\pi}{12}\right)\)

\(+\int_{0}^{\frac{\pi}{6}}[0] \mathrm{dx}+\int_{\pi / 6}^{\pi / 4}\left[2 \sin _{\downarrow} \mathrm{x}\right] \mathrm{dx} \)

\(= 24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right]+\frac{\pi}{4}-\frac{\pi}{6}\)

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

\(\alpha=12\)

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