Question:

Let $I = 24 \int_0^{\pi/12} \left(\left|\sin\left|4x - \frac{\pi}{12}\right|\right| + [2\sin x]\right) dx$ ... (i) Now $\left|4x - \frac{\pi}{12}\right| = \begin{cases} -4x + \frac{\pi}{12} & ; x < \frac{\pi}{48} \\ 4x - \frac{\pi}{12} & ; x \ge \frac{\pi}{48} \end{cases}$ Therefore from (i): $$I = 24\left[\int_0^{\pi/48} \sin\left(-4x + \frac{\pi}{12}\right)dx + \int_{\pi/48}^{\pi/12} \sin\left(4x - \frac{\pi}{12}\right)dx + \int_0^{\pi/6} [2\sin x]dx + \int_{\pi/6}^{\pi/12} [2\sin x]dx\right]$$ $$I = 24\left[\frac{1 - \cos\frac{\pi}{12}}{4} + \frac{-\cos\frac{\pi}{12} - 1}{4} - \frac{\pi}{4} + \frac{\pi}{6}\right]$$ $$I = 24\left(\frac{1}{2}\right) + \frac{\pi}{4} - \frac{\pi}{6}$$ $$I = 2\pi + 12 = 2\pi + \alpha$$ Therefore $\alpha = 12$