I want to calculate: $\oint_C\frac{1}{x^2 + y^2}dx + 2y\ dy$ where $C = \{(x,y)\mid x^2 + y^2 = 1\}$. And I want to do it without greens formula.
To calculate the line integral we must then Parameterize the circle $x^2 + y^2 = 1$ into the parameterized path function $f(t) = (\cos(t), \sin(t))$ for $0 \leq t\leq 2 \pi$. So the integral $\oint_C \frac{1}{x^2 + y^2} \, dx + 2y \, dy$ then becomes $\int_0^{2\pi} \frac{1}{\cos(\theta)^2 + \sin(\theta)^2}(-\sin(\theta)) + 2(\sin(\theta)) \cos(\theta)d\theta$ which becomes $\int_0^{2\pi} (-\sin(\theta)) + 2(\sin(\theta)) \cos(\theta)d\theta = \int_0^{2\pi} (\sin(\theta))(2\cos(\theta) -1)\,d\theta =0$?
Can anybody tell me if this is right/close to right? The zero makes me doubt it.