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Let $\omega \in \Omega^1(\Bbb R^2)$ with $\omega=e^x\cdot sin(y) dx + e^x \cdot cos(y) dy$

$(1)$Show that $\omega $ has got a vector potential and calculate it

$(2)$ Calculate the line integral $\int_\gamma\omega$ over the curve $\gamma:[0,1]\to \Bbb R^2 ,t\mapsto(t,arctan(t))$

For $(1)$: Would it suffices if $d\omega=0?$ if so, then how do I find $d\omega$ and if not, what do I have to do ?

$\int e^x \cdot sin(y) dx= e^x \cdot sin(y)+c_1(y)$

$\Rightarrow\ \frac{\partial (e^x \cdot sin(y)+c_1(y))}{\partial y}=e^x\cdot cos(y) =e^x \cdot cos(y)+c_1'(y)$

$\Rightarrow c_1(y)=k \in \Bbb R$

$\Rightarrow f(x,y)=e^xsin(y)$ is a vector potential

For $(2):$

$\int_\gamma\omega=\int_0^1df=f(\gamma(1))-f(\gamma(0))=f(1,\frac{\pi}{4})-f(0,0)=\frac{e\sqrt2}{2}$ Is that right or wrong ?

Thanks in advance.

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    To show that $\omega = df$ for some $f$ it's enough to show that $d\omega = 0$ (since $\Bbb R^2$ is simply connected). In this case, though, we actually want to find $f$, as you have done, so there's no gain in computing $d\omega$. – 2017-02-17
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    @Travis,OK, so for showing that $\omega$ has got a potential vector what have I to write down ? – 2017-02-17
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    You've basically done the calculation already: If $\omega = df$, then expanding gives $\omega = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$, so we're looking to solve the system $\frac{\partial f}{\partial x} = e^x \sin y, \frac{\partial f}{\partial y} = e^x \cos y$, which you've solved. – 2017-02-17

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You wrote: "$\int e^x \cdot sin(y) dx= e^x \cdot sin(y) (1+x)+c_1(y)$"

But this is false !

We have $\int e^x \cdot sin(y) dx= e^x \cdot sin(y)+c_1(y)$

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    Thank you haven't noticed it. – 2017-02-20