How would I use Green's Theorem to evaluate this Line Integral? If I have the closed loop $C$ (https://ibb.co/hK7xC5V) which is the union of $C_1,C_2,C_3$ and $C_4$ where
$$C_1=\sin(x)-2\pi,\text{where $x$ goes from 0 to $2\pi$}$$
$$C_2=-\sin(y)+2\pi,\text{where $y$ goes from $-2\pi$ to 0}$$
$$C_3=\sin(x),\text{where $x$ goes from $2\pi$ to 0}$$
$$C_4=-\sin(y),\text{where $y$ goes from 0 to $-2\pi$}$$
How would I evaluate $$\oint\limits_C\vec{F}\cdot d\vec{r}$$ using Green's Theorem?
(An example would also be appreciated with any field $\vec{F}$ which has nonzero $x$ and $y$ components)
 A: (1) Trivial case: If $\vec F$ is conservative, then $\oint_C\vec F\cdot\mathrm d\vec r=0$ for any path $C$.
(2) Simple, not-as-trivial case: If $\vec F(x,y)=\left(-\frac y2,\frac x2\right)$, then $\oint_C\vec F\cdot\mathrm d\vec r$ is simply the area of the region $D$ bounded by $C$, since
$$\frac{\partial\left(\frac x2\right)}{\partial x}-\frac{\partial\left(-\frac y2\right)}{\partial y}=\frac12+\frac12=1$$
$$\implies\oint_C\vec F\cdot\mathrm d\vec r=\iint_D\mathrm dA$$
Take $D'$ to be the region (shaded red below) to be
$$D'=\left\{(x,y)\mid0\le x\le2\pi\,\land\,\sin(x)-2\pi\le y\le\sin(x)\right\}$$
The area of blue-shaded regions is accounted for in $D'$ by virtue of symmetry, so that both $D$ and $D'$ share the same total area.

Then $\iint_D\mathrm dA$ is simply the area between two curves,
$$\iint_D\mathrm dA=\int_0^{2\pi}\int_{\sin(x)-2\pi}^{\sin(x)}\mathrm dy\,\mathrm dx=4\pi^2$$
(3.a) The more general case: Unfortunately, the symmetry argument wouldn't hold up for arbitrary vector field $\vec F$. In this case, I'm not sure there's any way to avoid splitting up $D$ into smaller, easy-to-integrate-over regions. Perhaps a clever change of coordinates can be employed, but nothing immediately comes to mind that would help in integrating over the entirety of $D$ in one fell swoop.
Consider partitioning $D$ along the lines $x=\frac\pi2$, $x=\frac{3\pi}2$, $y=\frac\pi2$, and $y=\frac{3\pi}2$. This splits up $D$ into $9$ smaller regions $D_1,D_2,\ldots,D_9$ whose boundaries are easy enough to deal with.

The red region highlighted here is the set
$$D_1=\left\{(x,y)\mid -\sin(y)\le x\le\frac\pi2\,\land\,-\frac\pi2\le y\le0\right\}\cup\left\{(x,y)\mid \sin^{-1}(y)\le x\le\frac\pi2\,\land\,0\le y\le1\right\}$$
and integrating over $D_1$ requires splitting up into two more integrals.
Suppose $\vec F(x,y)=(0,x^2)$ (just to make things relatively easy). Then this first region's contribution to the total line integral would be
$$\iint_{D_1}2x\,\mathrm dA=\int_{-\frac\pi2}^0\int_{-\sin(y)}^{\frac\pi2}x\,\mathrm dx\,\mathrm dy+\int_0^1\int_{-\sin^{-1}(y)}^{\frac\pi2}x\,\mathrm dx\,\mathrm dy=1-\frac\pi8+\frac{\pi^3}{16}$$
While manageable, this method forces one to painstakingly set up and compute $13$ total integrals.
(3.b) Better method for the general case: As Mark S. points out, you can indeed simplify the integral over $D$ by summing the five integrals over the square $[0,2\pi]\times[-2\pi,0]$ and the integrals over the "lobes" of $D$ that extend outside of the square (two of these lobes are highlighted in blue in the second case above), then subtract the four integrals over the lobes that dip into the square.

In terms of this sketch, you would integrate over the square and the lighter-shaded lobes, then subtract the integrals over the darker-shaded ones.
