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$\int_{0}^{3 π / 2} \sin [\frac{2 x}{π}] d x$ where $[.]$ denotes the greatest integer function is equal to

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$\frac{π}{2} (\sin 1 - \cos 1)$

$\frac{π}{2} (\sin 1 + \sin 2)$

$\frac{π}{2} (\sin 1 + \cos 1)$

$\frac{π}{2} (\sin 1 - \sin 2)$

answer is D.

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Detailed Solution

$\int_{0}^{3 π / 2} \sin [\frac{2 x}{π}] d x = \int_{0}^{π / 2} \sin [\frac{2 x}{π}] d x$

$+ \int_{π / 2}^{π} \sin [\frac{2 x}{π}] d x + \int_{π}^{3 π / 2} \sin [\frac{2 x}{π}] d x$

$= 0 + \sin 1. \int_{π / 2}^{π} d x + \sin 2 \int_{π}^{3 π / 2} d x$

$= \frac{π}{2} (\sin 1 + \sin 2)$

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