1. **Problem Statement:** Given the derivative of a function $f'(x) = (x^2 - 2) \sin(x)$ and the initial condition $f(-1) = 8$, find the original function $f(x)$. 2. **Formula and Approach:** To find $f(x)$ from $f'(x)$, we use the integral: $$f(x) = \int f'(x) \, dx + C$$ where $C$ is the constant of integration. 3. **Integration:** We need to integrate: $$\int (x^2 - 2) \sin(x) \, dx$$ This can be split as: $$\int x^2 \sin(x) \, dx - 2 \int \sin(x) \, dx$$ 4. **Integrate $\int x^2 \sin(x) \, dx$ using integration by parts:** Let $u = x^2$, $dv = \sin(x) dx$. Then $du = 2x dx$, $v = -\cos(x)$. So, $$\int x^2 \sin(x) \, dx = -x^2 \cos(x) + \int 2x \cos(x) \, dx$$ 5. **Integrate $\int 2x \cos(x) \, dx$ again by parts:** Let $u = 2x$, $dv = \cos(x) dx$. Then $du = 2 dx$, $v = \sin(x)$. So, $$\int 2x \cos(x) \, dx = 2x \sin(x) - \int 2 \sin(x) \, dx = 2x \sin(x) + 2 \cos(x) + C$$ 6. **Combine results:** $$\int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C$$ 7. **Integrate $\int \sin(x) \, dx$:** $$\int \sin(x) \, dx = -\cos(x) + C$$ 8. **Put it all together:** $$f(x) = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) - 2(-\cos(x)) + C = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x) + 2 \cos(x) + C$$ Simplify: $$f(x) = -x^2 \cos(x) + 2x \sin(x) + 4 \cos(x) + C$$ 9. **Use initial condition $f(-1) = 8$ to find $C$:** $$8 = -(-1)^2 \cos(-1) + 2(-1) \sin(-1) + 4 \cos(-1) + C$$ Since $\cos(-1) = \cos(1)$ and $\sin(-1) = -\sin(1)$, $$8 = -1 \cdot \cos(1) - 2 \cdot (-\sin(1)) + 4 \cos(1) + C = -\cos(1) + 2 \sin(1) + 4 \cos(1) + C$$ $$8 = 3 \cos(1) + 2 \sin(1) + C$$ So, $$C = 8 - 3 \cos(1) - 2 \sin(1)$$ 10. **Final answer:** $$f(x) = -x^2 \cos(x) + 2x \sin(x) + 4 \cos(x) + 8 - 3 \cos(1) - 2 \sin(1)$$