So I'm reading through Ordinary Differential Equations by Tenenbaum and Pollard, page 76, Example 9.5, and I find it bizarre. It says that $\displaystyle \cos y dx - (x \sin y - y^{2})dy = 0$ is exact on $\displaystyle \mathbb{R} \times \mathbb{R}$ which I readily believe. It then says we can integrate each part from $\displaystyle (0,0)$ to $\displaystyle (x,y)$, i.e. $\displaystyle \int_{0}^{x}\cos y dx + \int_{0}^{y}y^{2} dy$. The second integrand baffles me. The book hints that, because we're taking the integral with respect to $\displaystyle y$ at the point $\displaystyle (x_{0}, y_{0}) = (0,0)$ with $\displaystyle x$ fixed, we can therefore eliminate the $\displaystyle x \sin y$. But one: Why? Two: If that's so, can't we do the same with the other one and have it be $\displaystyle \int_{0}^{x}(\cos (0))dx = x$?