1. **State the problem:** We need to find the integral $$\int x^2 \cos(x) \, dx$$. 2. **Formula and method:** Use integration by parts, which states: $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = x^2 \implies du = 2x \, dx$$ $$dv = \cos(x) \, dx \implies v = \sin(x)$$ 4. **Apply integration by parts:** $$\int x^2 \cos(x) \, dx = x^2 \sin(x) - \int 2x \sin(x) \, dx$$ 5. **Second integration by parts for $$\int 2x \sin(x) \, dx$$:** Let $$u = 2x \implies du = 2 \, dx$$ $$dv = \sin(x) \, dx \implies v = -\cos(x)$$ 6. **Calculate:** $$\int 2x \sin(x) \, dx = -2x \cos(x) + \int 2 \cos(x) \, dx = -2x \cos(x) + 2 \sin(x) + C$$ 7. **Substitute back:** $$\int x^2 \cos(x) \, dx = x^2 \sin(x) - \left(-2x \cos(x) + 2 \sin(x)\right) + C$$ 8. **Simplify:** $$= x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) + C$$ **Final answer:** $$x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) + C$$ which corresponds to option (a).