What is the solution of these coupled differential equations ?
$\frac{dx}{dt}=-sin(2\pi(y-\alpha))sin^2(\pi (x-\beta)) $ eqn 1
$\frac{dy}{dt}= \;\; sin(2\pi(x-\beta))sin^2(\pi (y-\alpha))$ eqn 2
$\alpha$ and $\beta$ are constants. These are the equations of a vortex velocity field. Figure shows the result that I got when I tried to numerically follow the path for intial cordinates (x,y) = (6.5,6.25) and $(\alpha,\beta) = (5.76,6.01)$ with $dt = 1/12$ for 30 million iterations. 
$$\frac{dx}{dt}=-2 \sin(\pi(y-\alpha))\cos(\pi(y-\alpha)) \sin^2(\pi (x-\beta))\tag 1$$
$$\frac{dy}{dt}= 2\sin(\pi(x-\beta))\cos(\pi(x-\beta))\sin^2(\pi (y-\alpha)) \tag 2$$ $$\frac{dy}{dx}=-\frac{\cos(\pi(x-\beta))\sin(\pi (y-\alpha)) }{\cos(\pi(y-\alpha)) \sin(\pi (x-\beta))} \tag 3$$ The ODE $(3)$ is separable. $$\frac{\cos(\pi (y-\alpha)) }{ \sin(\pi(y-\alpha)) }dy=-\frac{\cos(\pi(x-\beta)) }{\sin(\pi (x-\beta))}dx$$ $$\ln|\sin(\pi(y-\alpha))| = -\ln|\sin(\pi(x-\beta))|+\text{constant}$$ General equation of the trajectories : $$\boxed{\sin(\pi(y-\alpha))\sin(\pi(x-\beta))=C}\tag 4$$ Each trajectory depends on $C$ which is defined from the initial point $(x_0\;,\;y_0)$
$$C=\sin(\pi(y_0-\alpha))\sin(\pi(x_0-\beta)) \tag 5$$
One put $\sin(\pi(y-\alpha))=\frac{C}{\sin(\pi(x-\beta))}$ from $(4)$ into $(1)$ in order to eliminate $y$. $$\frac{dx}{dt}=-2 \frac{C}{\sin(\pi(x-\beta))} \sqrt{1-\frac{C^2}{\sin^2(\pi(x-\beta))}} \sin^2(\pi (x-\beta))$$ $$\frac{dx}{dt}=-2C \sqrt{\sin^2(\pi(x-\beta))-C^2}$$ $$t=-\frac{1}{2C}\int_{x_0}^x \frac{d\chi}{\sqrt{\sin^2(\pi(\chi-\beta))-C^2}}$$ $$t=-\frac{1}{2C}\int_{x_0}^x \frac{d\chi}{\sqrt{1-\sin^2(\frac{\pi}{4}-\pi(\chi-\beta))-C^2}}$$ $$t=-\frac{1}{2C\sqrt{1-C^2}}\int_{x_0}^x \frac{d\chi}{\sqrt{1-\frac{1}{1-C^2}\sin^2(\frac{\pi}{4}-\pi(\chi-\beta))}} \tag 6$$ $$t=-\frac{1}{2\pi C\sqrt{1-C^2}}\int_{\pi(x_0-\frac{1}{4}+\beta)}^{\pi(x-\frac{1}{4}+\beta)} \frac{d\xi}{\sqrt{1-\frac{1}{1-C^2}\sin^2(\xi)}}$$
$$t(x) = -\frac{1}{2\pi C\sqrt{1-C^2}}\left( \text{F}\left(\pi(x-\frac{1}{4}+\beta)\:,\:\sqrt{\frac{1}{\sqrt{1-C^2}}}\right) -\text{F}\left(\pi(x_0-\frac{1}{4}+\beta)\:,\:\sqrt{\frac{1}{\sqrt{1-C^2}}}\right)\right)$$ $\text{F}$ denotes the Elliptic Integral of the first kind : http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html
For the inverse function $x(t)$ one can numerically integrate $(6)$ from $x_0$ up to the value of $x$ such as the value of $t$ be reached.
Proceed similarly from equations $(2)$ and $(4)$.
COMMENT :
The above result, analytical solution $(4)$ , is in contradiction with the numerical result from the OP computation. The analytical result is a periodic function which graphical representation is a closed curve.
My own numerical solving doesn't agree with the OP. On the figure below the curve from equation $(4)$ is drawn in red. The numerical calculus gives the curve drawn in black. They are undistinguishable one from the other. This confirmes that my analytical and numerical calculus are consistent.
I think that the increment of $t$ which was chosen $=1/12$ by the OP is much too large. The small numerical errors accumulate and cause the deviation from a closed curve to something like a spiral. I think that the increment of $t$ must be much smaller, for example $0.00001$ or less.
The above curve seems close to a circle because the chosen initial point $(x_0,y_0)$ is close to $(\alpha,\beta)$.
Another example below shows the result with initial point less close to $(\alpha,\beta)$ :