1. The problem is to find the indefinite integral $$\int x^2 \sin(2x) \, dx.$$\n\n2. We will use integration by parts, which states: $$\int u \, dv = uv - \int v \, du.$$\n\n3. Choose $$u = x^2$$ and $$dv = \sin(2x) \, dx.$$ Then, $$du = 2x \, dx$$ and $$v = -\frac{\cos(2x)}{2}$$ because $$\int \sin(2x) \, dx = -\frac{\cos(2x)}{2}.$$\n\n4. Apply integration by parts:\n$$\int x^2 \sin(2x) \, dx = u v - \int v \, du = -\frac{x^2 \cos(2x)}{2} - \int -\frac{\cos(2x)}{2} \cdot 2x \, dx = -\frac{x^2 \cos(2x)}{2} + \int x \cos(2x) \, dx.$$\n\n5. Now, solve $$\int x \cos(2x) \, dx$$ using integration by parts again. Let $$u = x$$ and $$dv = \cos(2x) \, dx,$$ so $$du = dx$$ and $$v = \frac{\sin(2x)}{2}.$$\n\n6. Applying integration by parts again:\n$$\int x \cos(2x) \, dx = u v - \int v \, du = \frac{x \sin(2x)}{2} - \int \frac{\sin(2x)}{2} \, dx = \frac{x \sin(2x)}{2} - \frac{1}{2} \int \sin(2x) \, dx.$$\n\n7. Evaluate $$\int \sin(2x) \, dx = -\frac{\cos(2x)}{2}.$$\n\n8. 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\n9. Finally, substitute this result back into step 4:\n$$\int x^2 \sin(2x) \, dx = -\frac{x^2 \cos(2x)}{2} + \frac{x \sin(2x)}{2} + \frac{\cos(2x)}{4} + C,$$ where $$C$$ is the constant of integration.\n\nThis is the final answer.