Specific example of integrating a 1-form over a curve I was given the following definition in my course but no corresponding examples:
Supppose $\gamma:[a,b]\rightarrow{M}$ is a smooth curve and $\omega$ a 1-form on $M$ (so $\omega:M\rightarrow{T^*M}$). Then we get a smooth function $t\mapsto\omega(\gamma(t))\gamma'(t)$ (which makes sense since $\omega(\gamma(t))\in{T_{\gamma(t)}^*}M$ and so must act on $\gamma'(t)\in{T_{\gamma(t)}}M$ to return a real number). Then define:
\begin{equation}
\int_{\gamma}\omega=\int_a^b\omega(\gamma(t))\gamma'(t)\mathrm{d}t
\end{equation}
where the right hand integral is the integral over $\mathbb{R}$ in the usual sense. 
Now I am trying to compute this in the case of $\gamma(t)=(\cos(t),\sin(t))$ being the unit circle and $\omega$ being the following 1-form:
\begin{equation}
\omega=\frac{x\,\mathrm{d}y-y\,\mathrm{d}x}{x^2+y^2}
\end{equation}
My problem is that I have no idea how to 'plug in' the curve $\gamma(t)$ into the 1-forms $\mathrm{d}x$ and $\mathrm{d}y$ as in the definition and get something I can understand/compute. This isn't an assessed problem but if you're uneasy doing the whole question for me then please can you explain how to generally approach these problems, perhaps through a similar example. Thanks in advance!
 A: We first note that $\omega(\gamma(t))$ (the $1$-form $\omega$ evaluated at $\gamma(t)$) is simply $\omega(\gamma(t))\gamma'(t)\,dt=\dfrac{x(t)dy-y(t)dx}{x(t)^2+y(t)^2}$. But $dx,dy$ map $\gamma'(t)=(x'(t),y'(t))$ to its components, so $\omega(\gamma(t))\gamma'(t)=\dfrac{x(t)y'(t)-y(t)x'(t)}{x(t)^2+y(t)^2}.$ Hence one regains the 'calc 3' notion of $(x,y)$ being replaced by $(x(t),y(t))$ and differentiated appropriately to get an integral over $t$.
A: This is actually just calc 3, substitute $x=\cos t$ $y=\sin t$ and use $dx=-\sin t dt
$ etc.It all simplifys to $\int dt$.
A: To elaborate on @Rene's answer for any future readers, the definition of integral you have been given is actually not required, and is a simple case of elementary calculations.
The unit circle has coordinates $x = \cos(t)$ and $y = \sin(t)$. By differentiating we find
$$
\frac{\mathrm{d}x}{\mathrm{d}t} = -\sin(t), \hspace{20pt} \frac{\mathrm{d}y}{\mathrm{d}t} = \cos(t)
$$
so that $\mathrm{d}x = -\sin(t)\,\mathrm{d}t$ and $\mathrm{d}y = \cos(t)\,\mathrm{d}t$. Thus,
$$
\omega = \frac{x\,\mathrm{d}y - y\,\mathrm{d}x}{x^{2} + y^{2}} = \frac{\cos(t)\cdot\big(\cos(t)\,\mathrm{d}t\big) - \sin(t)\cdot\big(-\sin(t)\,\mathrm{d}t\big)}{\cos^{2}(t) + \sin^{2}(t)},
$$
which simplifies to $\omega = \mathrm{d}t$. Now $\gamma$ is only a function on one variable, namely $t$, so that
$$
\int_{\gamma} \omega = \int_{\gamma} \mathrm{d}t = \int_{t=0}^{t=2\pi} \mathrm{d}t
$$
is an elementary integral and equals $2\pi$.
