Let's circulate!

First we substitute x=3cos(t), y=3sin(t) where t goes from 0 to $2\pi$ . Implemented the integrand writes as: $\frac{-6}{9-5\sin(t)^{2}}$

A table of listed integrals gives us the solution: $-arctan[\frac{2}{3}tan(t)] | _{0}^{2\pi}$

Last we must respect that tangent has 2 infinity locations over the integration region at $t=\frac{\pi}{2}$ and $t=\frac{3\pi}{2}$ . This makes it necessary to divide the integration into 4 quarter angles finally delivering the value -2 $\pi$ .

Consider the following contour integral $J=\oint_{\gamma} \frac{1}{z} dz,$ where $\gamma$ is the curve defined by the boundary of the ellipse, taken in counter-clockwise direction. Using Cauchy's integral formula , we obtain $J=2\pi i$ Now, we may rewrite the integral above as follows $J= \oint_{\gamma} \frac{1}{x+iy} (dx + i dy)= \oint_{\gamma} \frac{x dx +ydy}{x^2+y^2} + i \oint_{\gamma} \frac{x dy -y dx}{x^2+y^2}$ Equating the imaginary part of both sides, we obtain the result.

Remark: The above derivation shows that, the shape of the (closed rectifiable) curve over which the integral is taken does not matter as long as it includes the origin in its interior (As a simple example, try evaluating the integral over a unit square with the origin at the middle!). If the curve does not include the origin in its interior, the integral would be zero.

One can use parametric coordinates of an ellipse, ie acos(x),bsin(x).