1. We are asked to solve the integral $$\int x \sin(x) \cos(x) \, dx$$. 2. Use the double-angle identity for sine: $$\sin(2x) = 2 \sin(x) \cos(x)$$, so $$\sin(x) \cos(x) = \frac{\sin(2x)}{2}$$. 3. Substitute into 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. Use integration by parts for $$\int x \sin(2x) \, dx$$: Let $$u = x$$, so $$du = dx$$. Let $$dv = \sin(2x) dx$$, so $$v = -\frac{\cos(2x)}{2}$$. 5. Apply integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ 6. Calculate: $$\int x \sin(2x) \, dx = -\frac{x \cos(2x)}{2} + \frac{1}{2} \int \cos(2x) \, dx$$ 7. Integrate $$\int \cos(2x) \, dx$$: $$\int \cos(2x) \, dx = \frac{\sin(2x)}{2} + C$$ 8. Substitute back: $$\int x \sin(2x) \, dx = -\frac{x \cos(2x)}{2} + \frac{1}{2} \cdot \frac{\sin(2x)}{2} + C = -\frac{x \cos(2x)}{2} + \frac{\sin(2x)}{4} + C$$ 9. Recall the factor $$\frac{1}{2}$$ outside the integral: $$\int x \sin(x) \cos(x) \, dx = \frac{1}{2} \left(-\frac{x \cos(2x)}{2} + \frac{\sin(2x)}{4}\right) + C = -\frac{x \cos(2x)}{4} + \frac{\sin(2x)}{8} + C$$ 10. The answer matches option c): $$-\frac{x}{4} \cos(2x) + \frac{1}{8} \sin(2x) + C$$.