Line integral of a vector function involving sine and cosine I have line integral of a vector function: $\vec{F}=-e^{-x}\sin y\,\,\vec{i}+e^{-x}\cos y\,\,\vec{j}$
The path is a square on the $xy$ plane with vertices at $(0,0),(1,0),(1,1),(0,1)$
Of course it is a closed line integral, and I know the result should be zero. 
I am baffled how can you calculate $\sin y$ or $\cos y$ where $y$ is an actual coordinate point?!
 A: Actually, have you computed $\nabla \times \vec{F}$ and verified that is $\vec{0}$? Let's see: $$\nabla \times \vec{F} = \left|\begin{array}\vec{i} & \vec{j} & \vec{k} \\ \partial_x & \partial_y & \partial_z \\ e^{-x} \sin y & e^{-x} \cos y & 0 \end{array}\right| = (0,0,-e^{-x} \cos y - e^{-x} \cos y) \neq \vec{0}$$
This way, the integral does not have to be zero. In fact, using Green's Theorem, we get $$\begin{align} \int_C \vec{F} ~\mathrm{d}r &= \int_0^1 \int_0^1 -2e^{-x} \cos y ~\mathrm{d}x~ \mathrm{d}y \\ &= -2 \int_0^1e^{-x} \mathrm{d}x \int_0^1 \cos y~\mathrm{d}y \\ &= -2 (-e^{-1} - (-e^{-0}))(\sin 1 - \sin 0)  \\ &= -2\left(1-\frac{1}{e}\right) \sin 1 \\ &= 2\left(\frac{1}{e}-1\right) \sin 1\end{align}$$
And $\sin 1$ is a number, so we just leave at that.
A: Now that you've changed the function, we have
$$
(-e^{-x}\sin(y),e^{-x}\cos(y))=\nabla e^{-x}\sin(y)
$$
Thus,
$$
\begin{align}
\oint\nabla (e^{-x}\sin(y))\cdot\mathrm{d}\vec{r}
&=e^{-x}\sin(y)\Big]_{(0,0)}^{(0,0)}\\
&=0
\end{align}
$$
