Question: Let $\omega=xdx$.

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> What is $\int\_{S^1}\omega$?
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What I know:  

The final result should be $1$.  

We can take parametrization $p:[0,\pi]\rightarrow S^1$ with $p(t)=(\cos(t),\sin(t))$ and $Dp(t)=(\sin(t),\cos(t))$. Then
$$\int\_{S^1}\omega=\int\_{S^1}xdx=\int\_{p([0,\pi]}xdx=\int\_{[0,\pi]}p^\*xdx=\int\_{[0,\pi]}xdx(p(t)(D(p(t)).$$
Is this correct? How do I continue from here?

Answer: 
An interesting approach is to use Stokes, noting that $S^1$ is the boundary of the unit disk. Then $d\omega = dx\wedge dx=0$, so

$$\int\_{S^1} \omega = \int\_{B} d\omega=0$$

Alternatively, with the parametrization you gave the pullback of $\omega$ is

$$\cos t(\cos t)'=-\cos t\sin t=-\frac{\sin 2t}2$$

and this integrates to $0$ over $[0,2\pi)$.
Notice that the integral on $S^1$ and the integral on $S^1 \setminus \{(1,0)\}$ are equal, because $\{(1,0)\}$ is a null set. A convenient parametrization is $p : (0, 2\pi) \to S^1 \setminus \{(1,0)\}$ given by $p(t) = (\cos t, \sin t)$. Then, as you say,

$$\int \limits \_{S^1} \omega = \int \limits \_{S^1 \setminus \{(1,0)\}} \omega = \int \limits \_{(0,2\pi)} p^\* \omega = \int \limits \_{(0,2\pi)} \cos t \ \Bbb d (\cos t) = \int \limits \_0 ^{2\pi} \cos t (\cos t)' \ \Bbb d t = \frac {\cos^2 t} 2 \Bigg| \_0 ^{2\pi} = \frac 1 2 - \frac 1 2 = 0 .$$