1. **State the problem:** We want to evaluate the integral $$\int x \sqrt{16 - 3x^2} \, dx.$$\n\n2. **Identify the method:** This integral suggests a substitution because of the composite function inside the square root. Let’s use substitution where the inner function is set as a new variable.\n\n3. **Substitution:** Let $$u = 16 - 3x^2.$$ Then, differentiate both sides with respect to $$x$$:\n$$\frac{du}{dx} = -6x \implies du = -6x \, dx.$$\n\n4. **Rewrite the integral:** Notice that $$x \, dx = -\frac{1}{6} du$$. Substitute into the integral:\n$$\int x \sqrt{16 - 3x^2} \, dx = \int \sqrt{u} \left(-\frac{1}{6} du\right) = -\frac{1}{6} \int u^{1/2} du.$$\n\n5. **Integrate:** Use the power rule for integration $$\int u^n du = \frac{u^{n+1}}{n+1} + C$$ where $$n = \frac{1}{2}$$:\n$$-\frac{1}{6} \int u^{1/2} du = -\frac{1}{6} \cdot \frac{u^{3/2}}{\frac{3}{2}} + C = -\frac{1}{6} \cdot \frac{2}{3} u^{3/2} + C = -\frac{1}{9} u^{3/2} + C.$$\n\n6. **Back-substitute:** Replace $$u$$ with $$16 - 3x^2$$:\n$$-\frac{1}{9} (16 - 3x^2)^{3/2} + C.$$\n\n**Final answer:** $$\boxed{-\frac{1}{9} (16 - 3x^2)^{3/2} + C}.$$