Mathematics - Definite Integration Question with Solution | TestHub

Let $I=24{\int }_0^{\frac{\pi }{2}}\left(\sin \left|4x-\frac{\pi }{2}\right|+[2\sin x]\right)dx$ ...(i) Now $\left|4x-\frac{\pi }{12}\right|=\{\begin{array}{ll}-4x+\frac{\pi }{12}& ;x<\frac{\pi }{48}\\ 4x-\frac{\pi }{12}& ;x\ge \frac{\pi }{48}\end{array}$ $\therefore$ from $(\mathrm{i})$ $\begin{array}{rl}& I=24{\int }_0^{\frac{\pi }{48}}-\sin \left(4x-\frac{\pi }{12}\right)dx+{\int }_{\frac{\pi }{48}}^{\frac{\pi }{4}}\sin \left(4x-\frac{\pi }{12}\right)\\ & +{\int }_0^{\frac{\pi }{6}}[2\sin x]dx+{\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}[2\sin x]dx\\ & I=24\left[\frac{\left(1-\cos \frac{\pi }{12}\right)}{4}-\frac{\left(-\cos \frac{\pi }{12}-1\right)}{4}\right)+\frac{\pi }{4}-\frac{\pi }{6}\\ & I=24\left(\frac{1}{2}\right)+\frac{\pi }{4}-\frac{\pi }{6}\\ & I=2\pi +12=2\pi +\alpha \text{ (from above) }\\ & \therefore \alpha =12\end{array}$