Show that $\int_C \mathbf{A}\cdot d\mathbf{r}=\frac{5}{3}$, where $\mathbf{A}=(x^2-3y^2)\mathbf{i}+(y^2-2x^2)\mathbf{j}$ and $C$ is a closed curve in the $xy$-plane such that $$x=3\cos t, ~~ y=2\sin t, ~~~ 0\leq t\leq 2\pi.$$
\begin{align*} \int_C \mathbf{A}\cdot d\mathbf{r}&=\int_C(x^2-3y^2) dx + (y^2-2x^2) dy\\&=\int_C \left[-3\, \left( 9\, \cos^{2} \left( t \right) -12\, \sin^{2} \left( t \right) \right) \sin \left( t \right) +2\, \left( -18\, \cos^{2} \left( t \right) + 4\, \sin^{2} \left( t \right) \right) \cos \left( t \right)\right] dt\\&=\int_0^{2\pi}(-63\,\sin \left( t \right) \cos^{2} \left( t \right) -44\, \cos^{3} \left( t \right) +36\,\sin \left( t \right) +8\,\cos \left( t \right))dt\\&= 0 \end{align*}
Don't know where is the mistake. Please help me to solve.
Hint: Use Green's theorem to get the solution easily.