If
$$\gamma=(2\cos(t)+4\cos^2(t) , 2\sin(t)+4\sin(t)\cos(t))$$
and $C$ is a path defined by $\gamma$ when $\frac{-2\pi}{3} \le t\le \frac{4\pi}{3}$.
calculate:
$$\int_{C}\frac{-y\,dx+(x-1)\,dy}{(x-1)^2+y^2} $$
first of all, I was given a hint that $\gamma$ can be changed to polar coordinates, but it seems to me already in these coordinates. am I missing something?
second, for calculating the integral I thought to use that:
$$\int_{C}f\,dr=\int_{0}^1f(\gamma(t)\cdot\gamma'(t)\,dt$$
but with current parametrization it is too messy.
any ideas how to approach this question?
edit: I draw the function using matlab:
As I can judge it is the second type of line integral, while your sugested formula is used to calculate the first one.
For second type you have the following:
$$ \begin{cases} x = 2\cos t + 4\cos^2 t \\ y = 2\sin t + 4\sin t\cos t \end{cases}$$
so first of all you have to find:
$$\begin{cases} dx =(2\cos t +4\cos^2 t)'\,dt \\ dy = (2\sin t +4\sin t\cos t)'dt \end{cases}$$
After that you have to transder your integral to the defenite one with given limits, where $x$ will be your originally given $x$ parametric coordinate, $y$ your originally given $y$ parametric coordinate and therefore $dx$ and $dy$ are both ones come from derivatives I've written above.
After all that you will have to just simplify the integral and solve it as remarkable defenite integral with limits $[\frac{-2\pi}{3};\frac{4\pi}{3}]$.