help me to find the Line Integral Find  
 for  
on the curve counterclockwise around the unit circle C starting at the point (1,0).
I dont know the way to do that I tried many ways but I still could not get the right answer
 A: There's two possible methods you can use. Either you can use Green's Theorem (easy) or evaluate the line integral (harder). I'll do the line integral first:
First off, parametrise the unit circle by letting $\mathbf r = (\cos t ,\sin t )$,    $t \in [0, 2 \pi)$. Then $\mathbf r '= (-\sin t, \cos t)$ and using the formula 
$$\oint_C \mathbf F \cdot d\mathbf r = \int \mathbf F(\mathbf r(t)) \cdot \mathbf r ' (t) dt$$
We obtain: 
$$I = \oint_C \mathbf F \cdot d\mathbf r = \int_0^{2\pi} (6\sin t, \sin(\sin t))\cdot(-\sin t, \cos t) dt $$
$$=\int_0^{2\pi} -6\sin^2 t + \cos t \cdot \sin (\sin t) dt$$
The first half of this integral is easy enough to evaluate, the second can be evaluated by noticing $\frac{d}{dt} \cos(\sin t) = -\cos t \cdot \sin (\sin t)$. Thus the value of the integral is $I=-6\pi$.
Now by Green's Theorem. Green's Theorem tells us that:
$$I = \oint_C \mathbf F \cdot d\mathbf r = \iint_D \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} dxdy$$
Since $\frac{\partial F_2}{\partial x} = 0$, and $\frac{\partial F_1}{\partial y} = 6$, the result follows from the fact that the area of the unit circle is $\pi$.
