Please point out my stupid error.
edit: $C$ is positively oriented, I should add
First solution (correct one) comes by Green's Theorem. So the integral is just twice the area of the circle: $2\pi$.
Second solution by parameterization. Use $x=\cos(t), y=\sin(t)$ of course.
Then \begin{align} & \oint_C (2x-y)\,dx + (x+3y)\,dy \\[8pt] = {} & \int_0^{2\pi} (2\cos(t)-\sin(t))(-\sin(t))\,dt + (\cos(t)+3\sin(t))(\cos(t))\,dt \\[8pt] = {} & \int_0^{2\pi}(1+\cos(t)\sin(t))\,dt = 2\pi. \end{align}
and no problem now.
