Proof that winding number for certain vectorfield is an integer for closed $C^1$ curve Say I have the following vector field: $$\omega : \mathbb{R^2} \backslash \{0\} \to \mathbb{(R^2)}  
   ,(x,y) \mapsto (\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}). $$
I want to show that the winding number $n(c,0) = \frac{1}{2\pi} \int_c \omega $ around 0 for a smooth closed curve $c$  is an integer.
I know that my vector field is not conservative even though it fulfills $$\partial_{y}\omega_1 = \partial_{x}\omega_2.$$ I unsure how to proceed, but I feel like I have to use some special properties of the vector field, like the fact that it is the derivative of the Argument function, or the fact that it is the standard example of a closed but non-exact $1$-Form.
Further question: If I let $\omega$ be defined on an open, connected domain $D$ such that $\omega$ is exact on $D$, I want to show that $n(c,0) = \frac{1}{2\pi} \int_c \omega =0. $
 A: As mentioned in the comments, the idiomatic way to do this is via deformation: any closed curve in the punctured plane is homotopic to $t \mapsto (\cos(kt),\sin(kt))$ for some integer $k$; so it suffices to compute the integrals on these coverings of the circle and then use the fact that the integral of a closed form is homotopy invariant. Since you said you don't want to use topological tools, though, here's an elementary approach requiring nothing more than multivariable calculus.
Let $\gamma : [0,1] \to \mathbb R^2 \setminus\{0\}$ be a smooth parametrization of $c$. Since the curve is closed, we know $\gamma(0) = \gamma(1)$. 
Define the function $\phi(t) = \mathrm{Arg}(\gamma(0)) + \int_0^t \omega(\gamma(t)) \cdot \gamma'(t) dt$ so that $\int_c \omega = \phi(1) - \phi(0)$. The intuition here is that since $\omega = d \mathrm Arg$ locally, $\phi$ should give the argument of $\gamma$ (up to a multiple of $2 \pi$), and thus $\phi(1) - \phi(0)$ must be a multiple of $2\pi$ in order for the curve to close up.
In order to make this rigorous, I would proceed in two steps:


*

*Prove that $\gamma(t) = |\gamma(t)|(\cos \phi(t), \sin\phi(t))$ using calculus. (They agree at $t=0$ by definition, so it suffices to show their derivatives agree.)

*Thus prove that $\gamma(0) = \gamma(1)$ implies $\phi(1) = \phi(0) + 2 \pi n$ for some integer $n$.


For your further question, just apply the fundamental theorem of calculus and use the fact that the endpoints join up.
