선적분

The line integral is $\int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r}$ with $\mathbf{F} = (ye^x + \sin y)\,\hat{\boldsymbol{\imath}} + (e^x + \sin y + x \cos y)\,\hat{\boldsymbol{\jmath}}$. We are to show that it is independent of path. That is the case if and only if $\mathbf{F}$ is conservative. So let's look for a potential $\varphi$ for $\mathbf{F}$. That is, let's look for a function $\varphi$ that obeys

$$\begin{aligned} \frac{\partial \varphi}{\partial x}(x,y) &= ye^x + \sin y \\ \frac{\partial \varphi}{\partial y}(x,y) &= e^x + \sin y + x \cos y \end{aligned}$$

Integrating the first of these equations gives

$$\varphi(x,y) = ye^x + x\sin y + f(y)$$

Substituting this into the second equation gives

$$e^x + x\cos y + f'(y) = e^x + \sin y + x \cos y\qquad\text{or}\quad f'(y) = \sin y$$

which integrates to

$$f(y) = -\cos y + C$$

So $\mathbf{F}$ is indeed conservative with one potential being $\varphi(x,y) = ye^x + x\sin y -\cos y$ and the line integral is

$$\begin{aligned} \int_C (ye^x + \sin y)\,\mathrm{d}x + (e^x + \sin y + x \cos y)\,\mathrm{d}y &=\int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} =\varphi(x,y)\Big|^{(0,\pi/2)}_{(1,0)} \\ &=\Big[ye^x + x\sin y -\cos y\Big]^{(0,\pi/2)}_{(1,0)} \\ &=1+\frac{\pi}{2} \end{aligned}$$