If $24 \int_0^{\frac{\pi}{4}}\left(\sin \left|4 x-\frac{\pi}{12}\right|+[2 \sin x]\right) \mathrm{d} x=2 \pi+\alpha$, where $[\cdot]$ denotes the greatest integer function, then $\alpha$ is equal to _______.

Question

Aicharya · Accepted Answer

Let $I=24 \int_0^{\frac{\pi}{2}}\left(\sin \left|4 x-\frac{\pi}{2}\right|+[2 \sin x]\right) d x$ ...(i)
Now $\left|4 x-\frac{\pi}{12}\right|= \begin{cases}-4 x+\frac{\pi}{12} & ; x < \frac{\pi}{48} \\ 4 x-\frac{\pi}{12} & ; \quad x \geq \frac{\pi}{48}\end{cases}$
$\therefore$ from $(\mathrm{i})$
$\begin{aligned} & I=24 \int_0^{\frac{\pi}{48}}-\sin \left(4 x-\frac{\pi}{12}\right) d x+\int_{\frac{\pi}{48}}^{\frac{\pi}{4}} \sin \left(4 x-\frac{\pi}{12}\right) \\ & +\int_0^{\frac{\pi}{6}}[2 \sin x] d x+\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}[2 \sin x] d x \\ & I=24\left[\frac{\left(1-\cos \frac{\pi}{12}\right)}{4}-\frac{\left(-\cos \frac{\pi}{12}-1\right)}{4}\right)+\frac{\pi}{4}-\frac{\pi}{6} \\ & I=24\left(\frac{1}{2}\right)+\frac{\pi}{4}-\frac{\pi}{6} \\ & I=2 \pi+12=2 \pi+\alpha \text { (from above) } \\ & \therefore \quad \alpha=12\end{aligned}$

If $24 \int_0^{\frac{\pi}{4}}\left(\sin \left|4 x-\frac{\pi}{12}\right|+[2 \sin x]\right) \mathrm{d} x=2…

Solution