Question

${\int }_0^{3\pi /2}\sin \left[\frac{2x}{\pi }\right]\cdot dx$, where $[\cdot ]$ denotes the greatest integer function, is equal to

• A. $\frac{\pi }{2}(\sin 1+\cos 1)$
• B. $\frac{\pi }{2}(\sin 1+\sin 2)$
• C. $\frac{\pi }{2}(\sin 1-\cos 1)$
• D. $\frac{\pi }{2}(\sin \pi +\sin 2)$

Answer 1

The correct option is B $\frac{\pi }{2}(\sin 1+\sin 2)$

Let $I={\int }_0^{\frac{3\pi }{2}}\sin \left(\frac{2x}{\pi }\right)dx$

$={\int }_0^{\frac{\pi }{2}}\sin \left(\frac{2x}{\pi }\right)dx+{\int }_{\frac{\pi }{2}}^{\pi }\sin \left(\frac{2x}{\pi }\right)dx+{\int }_{\pi }^{\frac{3\pi }{2}}\sin \left(\frac{2x}{\pi }\right)dx$.
$=0+\sin 1{\int }_{\frac{\pi }{2}}^{\pi }dx+\sin 2{\int }_{\pi }^{\frac{3\pi }{2}}dx$
$=\frac{\pi }{2}(\sin 1+\sin 2)$