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

We need to evaluate: $$24\int_0^{\pi/4}\left(\sin\left|4x - \frac{\pi}{12}\right| + [2\sin x]\right)dx = 2\pi + \alpha$$

where $$[\cdot]$$ is the greatest integer function.

Part 1: $$\int_0^{\pi/4} \sin|4x - \pi/12|\,dx$$

$$4x - \pi/12 = 0$$ when $$x = \pi/48$$.

For $$0 \le x < \pi/48$$: $$|4x - \pi/12| = \pi/12 - 4x$$

For $$\pi/48 \le x \le \pi/4$$: $$|4x - \pi/12| = 4x - \pi/12$$

$$I_1 = \int_0^{\pi/48}\sin(\pi/12 - 4x)dx + \int_{\pi/48}^{\pi/4}\sin(4x - \pi/12)dx$$

$$= \left[\frac{\cos(\pi/12-4x)}{4}\right]_0^{\pi/48} + \left[-\frac{\cos(4x-\pi/12)}{4}\right]_{\pi/48}^{\pi/4}$$

$$= \frac{1}{4}[\cos 0 - \cos(\pi/12)] + \frac{1}{4}[-\cos(\pi - \pi/12) + \cos 0]$$

$$= \frac{1}{4}[1 - \cos(\pi/12)] + \frac{1}{4}[\cos(\pi/12) + 1]$$

$$= \frac{1}{4}[1 - \cos(\pi/12) + \cos(\pi/12) + 1] = \frac{2}{4} = \frac{1}{2}$$

Part 2: $$\int_0^{\pi/4}[2\sin x]\,dx$$

For $$x \in [0, \pi/4]$$: $$2\sin x$$ ranges from 0 to $$2\sin(\pi/4) = \sqrt{2} \approx 1.414$$.

$$[2\sin x] = 0$$ when $$2\sin x < 1$$, i.e., $$\sin x < 1/2$$, i.e., $$x < \pi/6$$.

$$[2\sin x] = 1$$ when $$1 \le 2\sin x < 2$$, i.e., $$\pi/6 \le x \le \pi/4$$.

$$I_2 = \int_0^{\pi/6} 0\,dx + \int_{\pi/6}^{\pi/4} 1\,dx = \pi/4 - \pi/6 = \pi/12$$

Combining:

$$24(I_1 + I_2) = 24\left(\frac{1}{2} + \frac{\pi}{12}\right) = 12 + 2\pi$$

So $$2\pi + \alpha = 2\pi + 12$$, giving $$\alpha = 12$$.

The answer is 12.