Use u-substitution, change of limits and the FTC to evaluate the integral. Do NOT use integral features of your calculator. \( \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} 6 \sin ^{2}(x) \cos (x) d x=[?] \)

Let \( u = \sin(x) \). Then, the derivative is \[ \frac{du}{dx} = \cos(x) \quad \Rightarrow \quad du = \cos(x)\,dx. \] Thus, the differential \( dx \) can be replaced using \( du = \cos(x)\,dx \) in the integral. The given integral is \[ \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 6 \sin^2(x) \cos(x)\,dx. \] Using the substitution \( u = \sin(x) \) and \( du = \cos(x)\,dx \), the integral becomes \[ 6\int_{u(\frac{\pi}{2})}^{u(\frac{3\pi}{2})} u^2\,du. \] Next, compute the new limits: \[ u\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) = 1, \quad u\left(\frac{3\pi}{2}\right) = \sin\left(\frac{3\pi}{2}\right) = -1. \] So, the limits of integration are from \( u = 1 \) to \( u = -1 \). This gives: \[ 6\int_{1}^{-1} u^2\,du. \] To handle the reversed limits, recall that \[ \int_a^b f(u)\,du = -\int_b^a f(u)\,du. \] Thus, the integral becomes: \[ 6\int_{1}^{-1} u^2\,du = -6\int_{-1}^{1} u^2\,du. \] Now, evaluate the integral: \[ \int_{-1}^{1} u^2\,du. \] The antiderivative of \( u^2 \) is \( \frac{u^3}{3} \), so \[ \int_{-1}^{1} u^2\,du = \left[\frac{u^3}{3}\right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}. \] Thus, we have: \[ -6\int_{-1}^{1} u^2\,du = -6\left(\frac{2}{3}\right) = -4. \] The value of the integral is: \[ \boxed{-4}. \]