Q106 / 200 · Manipal (MET) 2014 · Mathematics
$\int_0^{\frac{3 \pi}{2}} \sin \left[\frac{2 x}{\pi}\right] \cdot d x$, where $[\cdot]$ denotes the greatest integer function, is equal to
- A$\frac{\pi}{2}(\sin 1+\cos 1)$
- Correct answerB$\frac{\pi}{2}(\sin 1+\sin 2)$
- C$\frac{\pi}{2}(\sin 1-\cos 1)$
- D$\frac{\pi}{2}(\sin \pi+\sin 2)$
Solution
$\begin{aligned} & \int_0^{\frac{3 \pi}{2}} \sin \left(\frac{2 x}{\pi}\right) d x=\int_0^{\frac{\pi}{2}} \sin \left(\frac{2 x}{\pi}\right) d x+\int_{\frac{\pi}{2}}^\pi \sin \left(\frac{2 x}{\pi}\right) d x \\ & \quad+\int_\pi^{\frac{3 \pi}{2}} \sin \left(\frac{2 x}{\pi}\right) d x \\ & =0+\sin 1 \int_{\frac{\pi}{2}}^\pi d x+\sin 2 \int_\pi^{\frac{3 \pi}{2}} d x \\ & \end{aligned}$
$=\frac{\pi}{2}(\sin 1+\sin 2)$
Hence, the correct answer is option: 'B'.