Proof of the fact that one-form $\omega$ satisfies $\int_{S^1} \omega =0$ $\Rightarrow$ $\omega$ is exact. I have difficulty in understanding the proof of
$$ \mathrm{one-form} \ \omega \ \mathrm{satisfies} \ \int_{S^1} \omega =0 \Rightarrow \omega \  \mathrm{is \ exact}$$ where $S^1$ is an unit circle.
($\omega$ is exact $\underset{\mathrm{def}}\iff$ there exists $f\in C^\infty$ s.t. $df=\omega.$)
The proof is here.

Define $f : S^1 \to \mathbb R$ as $f(p)=\int_C \omega$ where $C$ is a curve from $(1,0)\in S^1$ to $p\in S^1$. Then, $df=\omega.$

I don't know why $df=\omega$ holds.
I think I have to calcurate $\int_C \omega$. And in calculating $\int_C \omega,$ I have to parametrize $C$ but I have no idea how I should parametrize.
I want you to give me any help.
 A: Here is what I got. Maybe you need to fill some details.
Consider the parametrization of $S^¹$ given by $\psi: (0,2\pi)\rightarrow S^1$
$$\psi(x)=e^{ix}$$
So, if $p=e^{iy}$, then $T_p S^1 = span\{i e^{iy}\}$
Since $\omega$ is an $1$-form, we can write it as $\omega(p) = \psi(p) dx|_p$, where $\psi \in C^\infty(S^1)$ and $dx|_p$ is the dual basis of $\{i e^{iy}\}$. Let $X=\lambda i e^{iy}\in T_p S^1$ and notice that $C$ can be parametrized by $\gamma(t)=e^{it}$, with $t\in[0,y]$. Consider the curve $\alpha(t)=e^{i(\lambda t+ y)}$ such that $\alpha(0)=p$ and $\alpha'(0)=X$.Then we have the following computation
$$df(X)=\frac{d}{dt}\mid_{t=0} f\circ \alpha(t)=\frac{d}{dt}\mid_{t=0}\int_{C_t}\omega,$$
where $C_t$ is the curve that goes from $(1,0)$ to $\alpha(t)$. Now we need to compute the integral
$$\int_{C_t}\omega = \int_{\psi^{-1}(C_t)}\psi^* \omega = \int_0^{\lambda t+y}\psi(e^{ix})dx$$
Taking the derivative at $t=0$ we get
$$\frac{d}{dt}\mid_{t=0}\int_0^{\lambda t+y}\psi(e^{ix})dx = \psi(e^{iy})\lambda = \omega(X)$$
as desired.
