how to find this line integral and what is its answer evaluate the line integral $$\int_C (xy^2 dy-x^2y dx), $$ taken in the counter-clockwise sense along the cardioid $$r= a(1+\cos\theta)$$
here putting the parametric form of cardioid $x=a(2\cos t-\cos2t), y= a(2\sin t-\sin2t) $ and taking $\theta$ , $0 $ to $2\pi $ tried  to solve but then it became complicated, so is there any other method and in this method please solve these as well ?

If $x=a(2\cos t-\cos 2t)$ and $y=a(2\sin t-\sin 2t)$, then
$$
xy=a^2\big(2\sin 2t-2\sin3t+\tfrac{1}{2}\sin 4t\big),
$$
$$
x^2-y^2=a^2\big(4\cos 2t-4\cos 3t+\cos4t\big),
$$
and
$$
x\,dx-y\,dy=a^2\big(-4\sin 2t+6\sin 3t-2\sin4t\big)\,dt.
$$
Hence
$$
xy^2\,dy-x^2y\,dx=a^4\big(2\sin 2t-2\sin3t+\tfrac{1}{2}\sin 4t\big)
\big(4\sin 2t-6\sin 3t+2\sin4t\big)\,dt,
$$
and
\begin{align}
\int_C xy^2\,dy-x^2y\,dx &=\int_0^{2\pi}a^4\big(2\sin 2t-2\sin3t+\tfrac{1}{2}\sin 4t\big)
\big(4\sin 2t-6\sin 3t+2\sin4t\big)\,dt\\ &=a^{4}\int_0^{2\pi}\big(8\sin^22t+12\sin^23t+\sin^44t\big)\,dt=\pi a^4 (8+12+1)=21\pi a^4.
\end{align}
