Let us have a closed and periodic curve $C$, in the unit sphere $\mathbb{S^2}$.(T understand better, visually the curve is like this: ~ , around the equator of the sphere). I have to calculate the geodesic curvature of $C$: $\int_{C}\mathcal{K}_g(s)ds$.
This is what I've done:
Let's define $\mathbb{S^2}=\Omega_1 \cup \Omega_2 \cup C$ where $\Omega_1$ is the upper part of the sphere and $\Omega_2$ the lower part of the sphere. So $\Omega_1$ and $\Omega_2$ are regular.
Using the global Gauss-Bonnet theorem, we know that:
$\int_C \mathcal{K}_g(s)ds+\iint_{\Omega_1}K dA=2\pi \mathcal{X}(\Omega_1)$
and
$-\int_C \mathcal{K}_g(s)ds+\iint_{\Omega_2}K dA=2\pi \mathcal{X}(\Omega_2)$
**Edit: On the one hand, as the curvature of the sphere is constant, I can get it out of the integral. On the other hand, the area of $\Omega_1$ and the area of $\Omega_2$ are equal. In addition, $\mathcal{X}(\Omega_1)=\mathcal{X}(\Omega_2)=1$ (because the surface is simple). So if I rest the second expression to the first one, I obtain $2\int_C \mathcal{K}_g(s)ds=0 \rightarrow \int_C \mathcal{K}_g(s)ds=0$.
**Edit2: I have correct some calculus mistakes.
Is this ok?