Integral Calculus
86459
\(\int_{0}^{\frac{3 \pi}{2}} \sin \left[\frac{2 x}{\pi}\right] \mathrm{dx}\), where [.] denotes the greatest integer function, is equal to
1 \(\frac{\pi}{2}(\sin 1+\cos 1)\)
2 \(\frac{\pi}{2}(\sin 1+\sin 2)\)
3 \(\frac{\pi}{2}(\sin 1-\cos 1)\)
4 \(\frac{\pi}{2}(\sin \pi+\sin 2)\)
Explanation:
(B) : Given,
\(\int_{0}^{\frac{3 \pi}{2}} \sin \left[\frac{2 x}{\pi}\right] d x\)
\(= \int_{0}^{\frac{\pi}{2}} \sin \left[\frac{2 x}{\pi}\right] d x+\int_{\frac{\pi}{2}}^{\pi} \sin \left[\frac{2 x}{\pi}\right] d x+\int_{\pi}^{\frac{3 \pi}{2}} \sin \left[\frac{2 x}{\pi}\right] d x\)
\(= 0+\sin 1 \int_{\frac{\pi}{2}}^{\pi} d x+\sin 2 \int_{\pi}^{\frac{3 \pi}{2}} \mathrm{dx}\)
\(= \sin 1\left(\pi-\frac{\pi}{2}\right)+\sin 2\left(\frac{3 \pi}{2}-\pi\right)\)
\(= \sin 1\left(\frac{\pi}{2}\right)+\sin 2\left(\frac{\pi}{2}\right)=\frac{\pi}{2}(\sin 1+\sin 2)\)