1. **State the problem:** We need to find the integral $$\int x^2 \sin x \, dx$$. 2. **Formula and method:** We will use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = x^2 \implies du = 2x \, dx$$ $$dv = \sin x \, dx \implies v = -\cos x$$ 4. **Apply integration by parts:** $$\int x^2 \sin x \, dx = -x^2 \cos x - \int -\cos x (2x) \, dx = -x^2 \cos x + 2 \int x \cos x \, dx$$ 5. **Integrate $$\int x \cos x \, dx$$ by parts again:** Let $$u = x \implies du = dx$$ $$dv = \cos x \, dx \implies v = \sin x$$ 6. **Apply integration by parts again:** $$\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C$$ 7. **Substitute back:** $$\int x^2 \sin x \, dx = -x^2 \cos x + 2(x \sin x + \cos x) + C = -x^2 \cos x + 2x \sin x + 2 \cos x + C$$ **Final answer:** $$\int x^2 \sin x \, dx = -x^2 \cos x + 2x \sin x + 2 \cos x + C$$