If 24 ∫_0^ π 4 (sin |4 x- π 12 |+[2 sin x] ) d x=2 π+α, where [·] denotes the greatest integer function, then α is equal to .

Let $\displaystyle I=24 \int_0^{\frac{\pi}{2}}\left(\sin \left|4 x-\frac{\pi}{2}\right|+[2 \sin x]\right) d x$I=24∫02π​​(sin​4x−2π​​+[2sinx])dx ...(i) Now $\left|4 x-\frac{\pi}{12}\right|= \begin{cases}-4 x+\frac{\pi}{12} & ; x < \frac{\pi}{48} \\ 4 x-\frac{\pi}{12} & ; \quad x \geq \frac{\pi}{48}\end{cases}$​4x−12π​​={−4x+12π​4x−12π​​;x<48π​;x≥48π​​ $\therefore$∴ from $(\mathrm{i})$(i) $\displaystyle \begin{aligned} & I=24 \int_0^{\frac{\pi}{48}}-\sin \left(4 x-\frac{\pi}{12}\right) d x+\int_{\frac{\pi}{48}}^{\frac{\pi}{4}} \sin \left(4 x-\frac{\pi}{12}\right) \\ & +\int_0^{\frac{\pi}{6}}[2 \sin x] d x+\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}[2 \sin x] d x \\ & I=24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right)+\frac{\pi}{4}-\frac{\pi}{6} \\ & I=24\left(\frac{1}{2}\right)+\frac{\pi}{4}-\frac{\pi}{6} \\ & I=2 \pi+12=2 \pi+\alpha \text { (from above) } \\ & \therefore \quad \alpha=12\end{aligned}$​I=24∫048π​​−sin(4x−12π​)dx+∫48π​4π​​sin(4x−12π​)+∫06π​​[2sinx]dx+∫6π​4π​​[2sinx]dxI=24[4(1−cos12π​)​−4(−cos12π​−1)​)+4π​−6π​I=24(21​)+4π​−6π​I=2π+12=2π+α (from above) ∴α=12​