$\vec{F}=\frac{x\hat{i}-y\hat{j}}{b^2x^2-a^2 y^2}$ be a planar vector field, Find $\oint_C\vec{F}.d\vec{r}$ For $a>0, b>0,$ let $\vec{F}=\frac{x\hat{i}-y\hat{j}}{b^2x^2-a^2 y^2}$ be a planar vector field. Let $$C=\{(x,y)\in\mathbb R^2|x^2+y^2=a^2+b^2\}$$ be a circle oriented anti-clockwise. Then $$\oint_C\vec{F}.d\vec{r}=?$$
My attempt
I parametrize the given curve $C$ such that $$x(t)=\sqrt{a^2+b^2}\cos t, y(t)=\sqrt{a^2+b^2}\sin t, 0\leq t < 2\pi$$
$$\oint_C\vec{F}.d\vec{r}=\int_0^{2\pi} \frac{xdx-ydy}{b^2x^2-a^2 y^2}=\int_0^{2\pi} \frac{-(a^2+b^2)\cos t \sin t-(a^2+b^2)\cos t \sin t}{b^2(a^2+b^2)\cos^2 t-a^2 (a^2+b^2)\sin^2 t}dt=\int_0^{2\pi} \frac{-2\cos t \sin t}{b^2\cos^2 t-a^2 \sin^2 t}dt$$
How do I complete the solution? Is there any short way to defeat the problem?
Based on the comment, I applied Stoke's theorem also
$$\oint_C\vec{F}.d\vec{r}=\int \int _{\overline {C^o}}\nabla \times \vec{F}.\hat{n}ds=\int \int _{\overline {C^o}}\frac{2xy(b^2-a^2)}{(b^2x^2-a^2y^2)^2}\hat{k}.\hat{n}ds=0.$$ Since, $\hat n $ lies in the $xy-plane$. So, $\hat n \perp \hat k.$
But the answer was given $\frac{2\pi}{ab}$

To start, I'll make the definition $R\equiv a^2+b^2$. So we are examining the circumference of a circle with radius $R$ centered at the origin. Our vector field is $\mathbf{F}=\frac{x\hat{\mathbf{i}}-y\hat{\mathbf{j}}}{b^2x^2-a^2y^2}$
We're first going to convert to polar coordinates, using the identity 
$$\begin{bmatrix}
\hat{\mathbf{i}}\\
\hat{\mathbf{j}}
\end{bmatrix} =\begin{bmatrix}
\cos( \theta ) & -\sin( \theta )\\
\sin( \theta ) & \cos( \theta )
\end{bmatrix}\begin{bmatrix}
\hat{\mathbf{r}}\\
\hat{\boldsymbol{\theta }}
\end{bmatrix}$$
We find, $$\mathbf{F}=\frac{(r\cos(\theta))(\cos(\theta)\hat{\mathbf{r}}-\sin(\theta)\hat{\boldsymbol{\theta }})-(r\sin(\theta)((\sin(\theta)\hat{\mathbf{r}}+\cos(\theta)\hat{\boldsymbol{\theta }})))}{b^2r^2\cos^2(\theta)-a^2r^2\sin^2(\theta)}$$
This can be simplified to
$$\mathbf{F} =\frac{r\cos^{2}( \theta )\hat{\mathbf{r}} -r\cos( \theta )\sin( \theta )\hat{\boldsymbol{\theta }} -r\sin^{2}( \theta )\hat{\mathbf{r}} -r\sin( \theta )\cos( \theta )\hat{\boldsymbol{\theta }}}{r^{2}\left( b^{2}\cos^{2}( \theta ) -a^{2}\sin^{2}( \theta )\right)}$$
Using some trig identities, and cancelling the spare $r$,
$$\mathbf{F} =\frac{\cos( 2\theta )\hat{\mathbf{r}} -\sin( 2\theta )\hat{\boldsymbol{\theta }}}{r^{2}\left( b^{2}\cos^{2}( \theta ) -a^{2}\sin^{2}( \theta )\right)}$$
Now we need to integrate $$\int\limits ^{2\pi }_{0}\int\limits ^{R}_{0}( \nabla \mathbf{\cdot F}) r\mathrm{d} r\mathrm{d} \theta $$
I'll let you take it from here. Remember to incorporate the scale factors when computing $\nabla \mathbf{\cdot F}$!
