Identify the integral and its limits.

Step 1: Identify the integral and its limits. The integral to evaluate is: $\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \sin^3 x \cos x \, dx$

Step 2: Use a substitution method. Let $u = \sin x$. Then, the differential $du$ is $du = \cos x \, dx$.

Step 3: Change the limits of integration based on the substitution. When $x = \frac{\pi}{2}$: $u = \sin\left(\frac{\pi}{2}\right) = 1$. When $x = \frac{3\pi}{2}$: $u = \sin\left(\frac{3\pi}{2}\right) = -1$.

Step 4: Rewrite the integral in terms of $u$ and evaluate. Substitute $u$ and $du$ into the integral with the new limits: $\int_{1}^{-1} u^3 \, du$ Now, integrate $u^3$ with respect to $u$: $\left[ \frac{u^{3+1}}{3+1} \right]_{1}^{-1} = \left[ \frac{u^4}{4} \right]_{1}^{-1}$

Step 5: Apply the limits of integration. $\frac{(-1)^4}{4} - \frac{(1)^4}{4}$ $= \frac{1}{4} - \frac{1}{4}$ $= 0$

The value of the definite integral is 0.

$\boxed{0}$