Q: Evaluate $\displaystyle \int_{\gamma} y^2 e^{2\sin{x}} \cos{x} dx + ye^{2\sin{x}} dy$ where $\displaystyle \gamma$ is the arc of the curve $\displaystyle y = \cos{x}$ from the point $\displaystyle (0,1)$ to the point $\displaystyle (\frac{\pi}{2},0)$.

A: We know that the integral is exact, meaning that we can choose any path from $\displaystyle (0,1)$ to $\displaystyle (\frac{\pi}{2},0)$. In the solutions they have gone through the origin, along the coordinate axes. So the path taken is from $\displaystyle (0,1)$ to $\displaystyle (0,0)$, and then to $\displaystyle (\frac{\pi}{2},0)$

The answers then state that:

The integral is then

$\displaystyle \int^0_1 ye^{0} dy + \int^{\frac{\pi}{2}}_0 0 dx$

Not sure how they get this? It looks like they've just subbed $\displaystyle x=0$ and $\displaystyle y=\frac{\pi}{2}$ into the first and second bit of the original integral?