1. **Problem statement:** Find the integral $$\int x \sin(x) \cos(x) \, dx$$. 2. **Formula and identities:** Use the double-angle identity for sine: $$\sin(2x) = 2 \sin(x) \cos(x)$$, so $$\sin(x) \cos(x) = \frac{\sin(2x)}{2}$$. 3. **Rewrite the integral:** $$\int x \sin(x) \cos(x) \, dx = \int x \frac{\sin(2x)}{2} \, dx = \frac{1}{2} \int x \sin(2x) \, dx$$. 4. **Integration by parts:** Let - $$u = x \implies du = dx$$ - $$dv = \sin(2x) dx \implies v = -\frac{\cos(2x)}{2}$$ 5. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ So, $$\frac{1}{2} \int x \sin(2x) \, dx = \frac{1}{2} \left( -\frac{x \cos(2x)}{2} + \int \frac{\cos(2x)}{2} \, dx \right) = -\frac{x \cos(2x)}{4} + \frac{1}{4} \int \cos(2x) \, dx$$. 6. **Integrate $$\int \cos(2x) dx$$:** $$\int \cos(2x) \, dx = \frac{\sin(2x)}{2} + C$$. 7. **Substitute back:** $$-\frac{x \cos(2x)}{4} + \frac{1}{4} \cdot \frac{\sin(2x)}{2} + C = -\frac{x \cos(2x)}{4} + \frac{\sin(2x)}{8} + C$$. **Final answer:** $$\boxed{-\frac{x}{4} \cos(2x) + \frac{1}{8} \sin(2x) + C}$$. This matches option (c).