1. **Problem:** Solve the integral $$\int x \sin^2(x) \, dx$$. 2. **Formula and rules:** Use the identity $$\sin^2(x) = \frac{1 - \cos(2x)}{2}$$ to simplify the integral. 3. **Rewrite the integral:** $$\int x \sin^2(x) \, dx = \int x \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int x (1 - \cos(2x)) \, dx = \frac{1}{2} \int x \, dx - \frac{1}{2} \int x \cos(2x) \, dx$$ 4. **Integrate the first part:** $$\frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4}$$ 5. **Integrate the second part using integration by parts:** Let $$u = x$$, $$dv = \cos(2x) dx$$. Then $$du = dx$$, $$v = \frac{\sin(2x)}{2}$$. Apply integration by parts formula: $$\int u \, dv = uv - \int v \, du$$ So, $$\int x \cos(2x) \, dx = x \cdot \frac{\sin(2x)}{2} - \int \frac{\sin(2x)}{2} \, dx = \frac{x \sin(2x)}{2} - \frac{1}{2} \int \sin(2x) \, dx$$ 6. **Integrate $$\int \sin(2x) dx$$:** $$\int \sin(2x) \, dx = -\frac{\cos(2x)}{2}$$ 7. **Substitute back:** $$\int x \cos(2x) \, dx = \frac{x \sin(2x)}{2} - \frac{1}{2} \left(-\frac{\cos(2x)}{2}\right) = \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4}$$ 8. **Combine all parts:** $$\int x \sin^2(x) \, dx = \frac{x^2}{4} - \frac{1}{2} \left( \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4} \right) + C = \frac{x^2}{4} - \frac{x \sin(2x)}{4} - \frac{\cos(2x)}{8} + C$$ 9. **Final answer:** $$\boxed{\frac{x^2}{4} - \frac{x}{4} \sin(2x) - \frac{1}{8} \cos(2x) + C}$$ This corresponds to option d).