To solve the integral \(\int_{\pi/2}^{3\pi/2} [\sin x] \, dx\), where \([·]\) denotes the greatest integer function, we first need to understand the behavior of the sine function over the interval \([\pi/2, 3\pi/2]\).
- The sine function \(\sin x\) has a range from 1 to -1, and it crosses zero at \(\pi\).
- On the interval \([\pi/2, \pi]\), \(\sin x\) decreases from 1 to 0.
- On the interval \([\pi, 3\pi/2]\), \(\sin x\) decreases from 0 to -1.
Now, we apply the greatest integer function \([\cdot]\) to \(\sin x\):
- On the interval \([\pi/2, \pi)\), \([\sin x] = 0\) because \(0 \leq \sin x \lt 1\).
- On the interval \([\pi, 3\pi/2)\), \([\sin x] = -1\) because \(-1 \leq \sin x \lt 0\).
This implies the given integral can be split into two intervals:
\[\int_{\pi/2}^{3\pi/2} [\sin x] \, dx = \int_{\pi/2}^{\pi} 0 \, dx + \int_{\pi}^{3\pi/2} (-1) \, dx\]Now, compute each part:
- For \(\int_{\pi/2}^{\pi} 0 \, dx\), the result is 0 as the integrand is zero over that interval.
- For \(\int_{\pi}^{3\pi/2} (-1) \, dx\), this simplifies to:
\[\int_{\pi}^{3\pi/2} (-1) \, dx = -1 \times \left[ x \right]_{\pi}^{3\pi/2} = -1 \times \left( \frac{3\pi}{2} - \pi \right) = -1 \times \frac{\pi}{2} = -\frac{\pi}{2}\]Thus, the value of the integral is \(-\frac{\pi}{2}\).
Therefore, the correct answer is \(- \frac{\pi}{2}\).