1. We are asked to find the integral of the function $x \sin^2(x)$.\n\n2. The integral is $\int x \sin^2(x) \, dx$. To solve this, we use the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$.\n\n3. Substitute the identity into the integral:\n$$\int x \sin^2(x) \, dx = \int x \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \int x (1 - \cos(2x)) \, dx$$\n\n4. Distribute $x$ inside the integral:\n$$= \frac{1}{2} \int (x - x \cos(2x)) \, dx = \frac{1}{2} \left( \int x \, dx - \int x \cos(2x) \, dx \right)$$\n\n5. Calculate the first integral:\n$$\int x \, dx = \frac{x^2}{2}$$\n\n6. For the second integral $\int x \cos(2x) \, dx$, use integration by parts. Let $u = x$, $dv = \cos(2x) dx$. Then $du = dx$, $v = \frac{\sin(2x)}{2}$.\n\n7. Apply integration by parts formula $\int u \, dv = uv - \int v \, du$:\n$$\int x \cos(2x) \, dx = x \cdot \frac{\sin(2x)}{2} - \int \frac{\sin(2x)}{2} \, dx = \frac{x \sin(2x)}{2} - \frac{1}{2} \int \sin(2x) \, dx$$\n\n8. Calculate $\int \sin(2x) \, dx$:\n$$\int \sin(2x) \, dx = -\frac{\cos(2x)}{2}$$\n\n9. Substitute back:\n$$\int x \cos(2x) \, dx = \frac{x \sin(2x)}{2} - \frac{1}{2} \left(-\frac{\cos(2x)}{2}\right) = \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4}$$\n\n10. Substitute all back into the original integral:\n$$\int x \sin^2(x) \, dx = \frac{1}{2} \left( \frac{x^2}{2} - \left( \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4} \right) \right) + C$$\n\n11. Simplify:\n$$= \frac{x^2}{4} - \frac{x \sin(2x)}{4} - \frac{\cos(2x)}{8} + C$$\n\nThis is the final answer.