To evaluate the integral \( \int_{0}^{\frac{3\pi}{2}} \sin\left( \left\lfloor \frac{2x}{\pi} \right\rfloor \right) dx \), we first look at the behavior of the function inside the integral.
The function \( \left\lfloor \frac{2x}{\pi} \right\rfloor \) can be evaluated as follows:
- For \( 0 \leq x < \frac{\pi}{2} \), \( \frac{2x}{\pi} \) ranges from 0 to 1, hence \( \left\lfloor \frac{2x}{\pi} \right\rfloor = 0 \).
- For \( \frac{\pi}{2} \leq x < \pi \), \( \frac{2x}{\pi} \) ranges from 1 to 2, thus \( \left\lfloor \frac{2x}{\pi} \right\rfloor = 1 \).
- For \( \pi \leq x < \frac{3\pi}{2} \), \( \frac{2x}{\pi} \) ranges from 2 to 3, so \( \left\lfloor \frac{2x}{\pi} \right\rfloor = 2 \).
Therefore, we can split the integral into three parts based on these intervals:
- \(\int_{0}^{\frac{\pi}{2}} \sin(0) \, dx = \int_{0}^{\frac{\pi}{2}} 0 \, dx = 0\)
- \(\int_{\frac{\pi}{2}}^{\pi} \sin(1) \, dx = \sin(1) \left[ x \right]_{\frac{\pi}{2}}^{\pi} = \sin(1) \left( \pi - \frac{\pi}{2} \right) = \frac{\pi}{2} \sin(1)\)
- \(\int_{\pi}^{\frac{3\pi}{2}} \sin(2) \, dx = \sin(2) \left[ x \right]_{\pi}^{\frac{3\pi}{2}} = \sin(2) \left( \frac{3\pi}{2} - \pi \right) = \frac{\pi}{2} \sin(2)\)
Adding these results together, we get:
\[ \int_{0}^{\frac{3\pi}{2}} \sin\left( \left\lfloor \frac{2x}{\pi} \right\rfloor \right) \, dx = 0 + \frac{\pi}{2} \sin(1) + \frac{\pi}{2} \sin(2) = \frac{\pi}{2} (\sin 1 + \sin 2) \]
Therefore, the answer is \( \frac{\pi}{2}(\sin 1 + \sin 2) \), which matches the provided correct option.