line integral - what kind? I need to calculate $\int_\Gamma F(x) \, dx$ from $(0,1)$ to $(-1,0)$ of the unit circle.
$$F(x) = (x,y)$$
Now the answer is:

But I don't understand what they did. Why $\Gamma(t) = (\cos t, \sin t)$ & $t: \pi/2 \mapsto \pi$?
And why the calculate I like this?
I want to understand this kind of integration and I'd like to get some help. Thanks for any kind of help. 
 A: It looks like you've copied a few things down wrong.
The unit circle is parametrised by $\Gamma(t) = (\cos t, \sin t)$, where $0 \le t \le 2\pi$. This parametrisation starts at $(1,0)$ and runs, anti-clockwise, all the way around and back to $(1,0)$. You can substitute in some $t$-values and you'll see that $\Gamma(\frac{\pi}{2})=(0,1)$ and $\Gamma(\pi) = (-1,0)$. So, taking $\frac{\pi}{2} \le t \le \pi$ runs around the circle, anti-clockwise, from $(0,1)$ to $(-1,0)$.
For the line integral of a force you find the integral
$$\int {\bf F}({\bf x}) \cdot \mathrm{d}{\bf x}$$
If we make the substitution ${\bf x} = \Gamma(t)=(\cos t, \sin t)$ then ${\bf F}({\bf x}) = (x,y) = (\cos t, \sin t)$. The differential $\mathrm{d}{\bf x}$ changes as well. We have
$$\mathrm{d}{\bf x} = \mathrm{d}\left[\Gamma(t)\right] = \frac{\mathrm{d}\Gamma}{\mathrm{d}t}\mathrm{d}t$$
Since $\Gamma(t) = (\cos t, \sin t)$ we have $\mathrm{d}\Gamma/\mathrm{d}t = (-\sin t, \cos t)$. That gives
$$\int_{\pi/2}^{\pi} (\cos t, \sin t) \cdot (-\sin t, \cos t)~\mathrm{d}t$$
A: Your circle has radius $r=1$. For any circle with radius $a$ and centered at the origin we have the parametrization 

$$ (a\cos(t),a\sin(t)). $$

A: Recall from trigonometry that the point $(\cos t,\sin t)$ moves around the circle of unit radius centered at $(0,0)$, and has the following values:
$$
\begin{array}{c|c}
t & (\cos t,\sin t) \\
\hline
0 & (1,0) \\
\pi/2 & (0,1) \\
\pi & (-1,0) \\
3\pi/2 & (0,-1) \\
2\pi & (1,0) \\
\hline
\end{array}
$$
Then it has gone all the way around the circle and starts over.  This explains why $t$ goes from $\pi/2$ to $\pi$.
Then, differentiating, we get
\begin{align}
x & = (\cos t,\sin t) \\[8pt]
dx & = (-\sin t,\cos t)\,dt
\end{align}
and $F(x)$ is the identity function, so we're integrating $x\cdot dx$.
So $F(x)\cdot dx = (\cos t, \sin t)\cdot(-\sin t,\cos t)\,dt$.
