1. **State the problem:** Evaluate the integral $$\int_0^1 (x^2 \sin y - x \cos y) \, dx$$ where $y$ is treated as a constant with respect to $x$. 2. **Recall the integral rules:** Since $y$ is constant, $\sin y$ and $\cos y$ are constants during integration with respect to $x$. 3. **Split the integral:** $$\int_0^1 (x^2 \sin y - x \cos y) \, dx = \sin y \int_0^1 x^2 \, dx - \cos y \int_0^1 x \, dx$$ 4. **Evaluate each integral:** - $$\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}$$ - $$\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$$ 5. **Substitute back:** $$\sin y \cdot \frac{1}{3} - \cos y \cdot \frac{1}{2} = \frac{\sin y}{3} - \frac{\cos y}{2}$$ 6. **Rewrite with common denominator 6:** $$\frac{2 \sin y}{6} - \frac{3 \cos y}{6} = \frac{2 \sin y - 3 \cos y}{6}$$ 7. **Add constant of integration $C$ for indefinite integral:** $$\frac{2 \sin y - 3 \cos y}{6} + C$$ **Final answer:** $$\frac{2 \sin y - 3 \cos y}{6} + C$$ which corresponds to option C.