Calculate $$\int_C \frac{2xy^2dx-2yx^2dy}{x^2+y^2},$$ where $C$ is the ellipse $3x^2 +5y^2 = 1$ taken in the positive direction.
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my trial :
$~~~~~~~~~~~~~~~~~$The Elipse
$x(t) = \frac{\cos(t)}{\sqrt{3}}$ $~y(t) = \frac{\sin(t)}{\sqrt{5}}$
$t \in [0,2\pi]$
$ dx = \frac{-\sin(t)}{\sqrt{3}}~dt$
$ dy = \frac{\cos(t)}{\sqrt{5}}~dt$
$ 2xy^2~dx = \frac{2\cos(t)}{\sqrt{3}}\frac{\sin(t)^2}{5}\frac{-\sin(t)}{\sqrt{3}}~dt = \frac{-2\cos(t)\sin(t)^2\sin(t)}{15}~dt$
$ -2yx^2~dy = \frac{-2\sin(t)}{\sqrt{5}}\frac{\cos(t)^2}{3}\frac{\cos(t)}{\sqrt{5}}~dt = \frac{-2\cos(t)\cos(t)^2\sin(t)}{15}~dt$
$ x^2 + y^2 = \frac{\cos^2(t)}{3} + \frac{\sin^2(t)}{5} = \frac{3+2\cos^2(t)}{15}$
so $~\int_C \frac{2xy^2dx-2yx^2dy}{x^2+y^2} = \int_0^{2\pi}\frac{-2\sin(t)\cos(t)}{3+2\cos^2(t)}~dt = \frac{ln(3+2\cos^2(t))}{2}|_0^{2\pi} = 0$