1. **State the problem:** We need to evaluate the definite integral $$\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \cos\left(\frac{3\pi}{2} x\right) \, dx.$$\n\n2. **Recall the formula:** The integral of cosine is given by $$\int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C,$$ where $a$ is a constant.\n\n3. **Apply the formula:** Here, $a = \frac{3\pi}{2}$. So,\n$$\int \cos\left(\frac{3\pi}{2} x\right) dx = \frac{1}{\frac{3\pi}{2}} \sin\left(\frac{3\pi}{2} x\right) + C = \frac{2}{3\pi} \sin\left(\frac{3\pi}{2} x\right) + C.$$\n\n4. **Evaluate the definite integral:** Substitute the limits $x=\frac{3\pi}{4}$ and $x=\frac{\pi}{2}$:\n$$\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \cos\left(\frac{3\pi}{2} x\right) dx = \left[ \frac{2}{3\pi} \sin\left(\frac{3\pi}{2} x\right) \right]_{\frac{\pi}{2}}^{\frac{3\pi}{4}} = \frac{2}{3\pi} \left( \sin\left(\frac{3\pi}{2} \cdot \frac{3\pi}{4}\right) - \sin\left(\frac{3\pi}{2} \cdot \frac{\pi}{2}\right) \right).$$\n\n5. **Simplify inside the sine functions:**\nCalculate the arguments:\n$$\frac{3\pi}{2} \cdot \frac{3\pi}{4} = \frac{9\pi^2}{8}, \quad \frac{3\pi}{2} \cdot \frac{\pi}{2} = \frac{3\pi^2}{4}.$$\n\n6. **Evaluate the sine values:** Since $\sin$ is periodic with period $2\pi$, and these arguments are multiples of $\pi^2$, which is not a multiple of $\pi$, the sine values are not standard angles. We leave the answer in exact form:\n$$\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \cos\left(\frac{3\pi}{2} x\right) dx = \frac{2}{3\pi} \left( \sin\left(\frac{9\pi^2}{8}\right) - \sin\left(\frac{3\pi^2}{4}\right) \right).$$\n\n**Final answer:**\n$$\boxed{\frac{2}{3\pi} \left( \sin\left(\frac{9\pi^2}{8}\right) - \sin\left(\frac{3\pi^2}{4}\right) \right)}.$$