If $24\displaystyle\int_0^{\pi/6}\left(\sin\!\left|4x-\frac{\pi}{12}\right|+\left[2\sin x\right]\right)dx = 2\pi+\alpha$, where $[\cdot]$ denotes the greatest integer function, then $\alpha$ is equal to ____.
Step-by-Step Solution
Key Concept: Split $|4x-\pi/12|$ at $x = \pi/48$ and compute each piece; for $[2\sin x]$ on $(0,\pi/6)$ observe $2\sin x \in [0,1)$ so $[2\sin x] = 0$.
On $[0,\pi/6]$, $2\sin x \in [0,1)$, so $[2\sin x] = 0$.
$|4x-\pi/12|$: sign change at $x=\pi/48$.
$$I = 24\left[\int_0^{\pi/48}\!\sin\!\left(\frac{\pi}{12}-4x\right)dx + \int_{\pi/48}^{\pi/4}\!\sin\!\left(4x-\frac{\pi}{12}\right)dx\right]$$
Evaluating each piece:
$$I = 24\left[\frac{1-\cos\frac{\pi}{12}}{4} + \frac{\cos\frac{\pi}{12}+1}{4}\right] + 24\left(\frac{\pi}{4}-\frac{\pi}{6}\right) = 24\cdot\frac{1}{2} + 24\cdot\frac{\pi}{12} = 12+2\pi.$$
So $2\pi + \alpha = 2\pi + 12 \Rightarrow \alpha = 12$.
Correct Answer: 12