Let $C$ be the unit circle with positive orientation.
Find $$\int\limits_C e^{x^2-y^2}(\cos(2xy)dx+\sin(2xy)dy)$$
As I can see the differential form $e^{x^2-y^2}\cos(2xy)dx+e^{x^2-y^2}\sin(2xy)dy$ is not a closed form so it is also not exact. So, I tried to apply Green's theorem to get: $$\iint\limits_B 4xe^{x^2 - y^2}\sin(2 x y) + 4ye^{x^2-y^2}\cos(2 x y)dxdy$$ With $B$ being the open unit disk. The thing is that a parametrization for the above will end up with expressions like $\sin(2r\cos\theta r\sin\theta)$ and things like that which I couldn't solve.
Is there another way to solve this?
Do not be frightened by its appearance. Notice that for each given $x$ the functions
$$ y \mapsto 4x e^{x^2-y^2}\sin(2xy) \quad \text{and} \quad y \mapsto 4y e^{x^2-y^2}\cos(2xy) $$
are odd. Since your domain of integration $B$ is symmetric about the $y$-reflection, it follows that the double integral is simply zero.
Alternatively, if you know a bit about complex analysis then the original line integral can be written as the following contour integral
$$ \operatorname{Re}\left(\int_C \frac{e^{z^2}}{z^2} \, dz\right) $$
Now by the residue theorem, this integral can be computed to be zero.