Killing singularities of differential forms via multiplying with a function which is 1 on the boundary of the domain of integration. The most famous differential form is probably
$$\textrm{d}\theta =  \frac{-y}{x^2+y^2}\textrm{d}x + \frac{x}{x^2+y^2}\textrm{d}y$$
Warning:  $\textrm{d}\theta$, despite appearances, is not the exterior derivative of any function $\theta : \mathbb{R}^2 \to \mathbb{R}$.  It is locally of such a form however.  On any simply connected open set $U$ of $\mathbb{R}^2-\{(0,0)\}$ one can define a function $\theta_U : U \to \mathbb{R}$ whose differential agrees with $\textrm{d}\theta$ on $U$.
There are several ways to integrate this around the unit circle:

*

*Parameterize the unit circle and integrate in the parameter space.

*Cover the unit circle with simply connected open sets where the form is exact (for instance $\textrm{d}\theta = \textrm{d} \arctan(\frac{y}{x}) $ when $x>0$), chop the circle into pieces whose endpoints are in the overlaps of the open sets, and use the fundamental theorem of calculus for line integrals on each segment.

There is a third way which doesn't seem to be discussed as often:  multiply this form by a smooth function which kills the singularity at the origin, but does not alter the form on the unit circle, and then apply Stoke's theorem.  Here is that calculation in detail:
$$
\begin{align}
\int_{S^1}  \frac{-y}{x^2+y^2}\textrm{d}x + \frac{x}{x^2+y^2}\textrm{d}y & = \int_{S^1}  (x^2+y^2)\left(\frac{-y}{x^2+y^2}\textrm{d}x + \frac{x}{x^2+y^2}\textrm{d}y\right)\\
&=\int_{S^1} -y \textrm{d}x + x\textrm{dy}\\
&= \int_{D(0,1)} \textrm{d}\left( -y \textrm{d}x + x\textrm{dy} \right)\\
&= \int_{D(0,1)} 2 \textrm{d}x \wedge \textrm{d}y\\
&= 2\pi
\end{align}
$$
Here the first equality is justified by the fact that $x^2+y^2 = 1$ on $S^1$, the third equality is Stoke's theorem, and the last equality is true because the area of the unit disk is $\pi$.
It is interesting that this argument relates the circumference of the circle to its area via Stoke's theorem.  A slight variant of this calculation can prove that $C = 2A/r$ for any circle of radius $r$.
This trick seems widely applicable:  if you have a form which is defined off of a singular set, you can integrate along a boundary by first multiplying the form by a function which is 1 on the boundary and kills the singularity, and then applying Stoke's theorem to the new "singularity free" form.  It has the feeling of being an important trick to me, one which can be used over and over again in many diverse situations.
My question:  where has this trick been used?  I would like a "big list" of applications.
 A: Consider the differential equation $Mdx+Ndy=0$ where $M,N$ have domain of definition $U \subseteq \mathbb{R}^2$. If $\partial_y M = \partial_x N$ on $U$ and $U$ is simply connected then there exists $F: U \rightarrow \mathbb{R}$ such that $dF = Mdx+Ndy$ and hence the solution set of $F(x,y)=c$ constitutes a solution of the exact differential equation $Mdx+Ndy=0$. We say $I$ is an integrating factor of the inexact differential equation $Mdx+Ndy=0$ if the corresponding differential equation $IMdx+INdy=0$ is exact. That is, there exists $\phi$ for which $d \phi = IMdx+INdy$. Upon writing this it is clear my initial hunch seems completely off-base.
In fact, this example is almost exactly backwards of the integrating factor method! The initial form
$$ \omega = \frac{-ydx+xdy}{x^2+y^2} $$
is closed on its domain of definition as follows from the calculus below:
$$ \frac{\partial}{\partial y} \left[ \frac{-y}{x^2+y^2} \right]  = \frac{y^2-x^2}{(x^2+y^2)^2} \qquad \& \qquad \frac{\partial}{\partial x} \left[ \frac{x}{x^2+y^2} \right]  = \frac{y^2-x^2}{(x^2+y^2)^2}. $$
In fact, multiplying by $x^2+y^2$ spoils the closed condition. Note, multiplication by $x^2+y^2$ in the OP's example yielded $-ydx+xdy$ and $d(-ydx+xdy) = 2dx \wedge dy$ hence $-ydx+xdy$ is not closed which indicates $-ydx+xdy$ is not exact.
Instead, I think the way to understand the calculation is more rooted in the theory of residues.
In complex notation $z = x+iy$ and $dz = dx+idy$ so formally $x = \frac{1}{2}(z+\bar{z})$ and $y = \frac{1}{2i}(z- \bar{z})$ and $dx = \frac{1}{2}(dz+d\bar{z})$ and $dy = \frac{1}{2i}(dz- d\bar{z})$
\begin{align} 
\frac{-ydx+xdy}{x^2+y^2} &= \frac{-\frac{1}{2i}(z- \bar{z})\frac{1}{2}(dz+d\bar{z})+\frac{1}{2}(z+\bar{z})\frac{1}{2i}(dz- d\bar{z})}{z\bar{z}} \\
&= \frac{-1}{4iz\bar{z}}\left(zdz-\bar{z}dz+zd\bar{z}-\bar{z}d\bar{z} 
-(zdz+\bar{z}dz-zd\bar{z}-\bar{z}d\bar{z}) \right) \\
&= \frac{-1}{4iz\bar{z}}\left(-2\bar{z}dz+2zd\bar{z}\right) \\
&= \frac{\bar{z}dz}{2iz\bar{z}} -  \frac{zd\bar{z}}{2iz\bar{z}} \\
&= \frac{1}{2i}\left[\frac{dz}{z} -  \frac{d\bar{z}}{\bar{z}}\right] \\
\end{align}
Then integration around a CCW picks up $2\pi i$ from the $\frac{dz}{z}$ term and $-2\pi i$ from the $\frac{d\bar{z}}{\bar{z}}$ term. Hence $\frac{1}{2i}(2\pi i-(-2\pi i)) = 2\pi$.
Can we come up with an example which moves beyond just a single singularity ? It's a bit silly, but, I'd try:
\begin{align} \gamma &= \omega+ \omega_1  \\
&= \frac{-ydx+xdy}{x^2+y^2} + \frac{-(y-y_1)dx+(x-x_1)dy}{(x-x_1)^2+(y-y_1)^2} \\
&= \frac{1}{2i} \left( \frac{1}{z} + \frac{1}{z-z_1}\right)dz - 
\frac{1}{2i} \left( \frac{1}{\bar{z}} + \frac{1}{\bar{z}-\bar{z}_1}\right)d\bar{z}.
\end{align}
This form is closed except at its singularities $z=0$ and $z = z_1$. This gives us the freedom to adjust line integrals by deforming curves until they encircle the appropriate singularities. Suppose $C$ is a CCW-curve which encloses both $z=0$ and $z=z_1$. Then, I propose we can deform $C$ into CCW-oriented circles $C_0$ and $C_1$ of radii $R_0$ and $R_1$ respective. Then on each circle I can use the OP's trick: multiply by $\frac{1}{R_0^2}z\bar{z} = 1$ for $C_0$ and multiply by $\frac{1}{R_1^2}(z-z_1)(\bar{z}-\bar{z}_1) = 1$ for $C_1$. Then,
$$ \frac{1}{R_0^2}z\bar{z}\gamma = \frac{1}{2iR_0^2} \left( \bar{z}dz-zd\bar{z}+\frac{z\bar{z}dz}{z-z_1}-\frac{z\bar{z}d\bar{z}}{\bar{z}-\bar{z}_1} \right)$$
likewise,
$$ \small\frac{1}{R_1^2}(z-z_1)(\bar{z}-\bar{z}_1)\gamma = \frac{1}{2iR_1^2} \left( (\bar{z}-\bar{z}_1)dz-(z-z_1)d\bar{z}+\frac{(z-z_1)(\bar{z}-\bar{z}_1)dz}{z}-\frac{(z-z_1)(\bar{z}-\bar{z}_1)d\bar{z}}{\bar{z}} \right)$$
I suspect the non-singular (on $C_0$ and $C_1$ respective) terms $\frac{z\bar{z}dz}{z-z_1}-\frac{z\bar{z}d\bar{z}}{\bar{z}-\bar{z}_1}$ and $\frac{(z-z_1)(\bar{z}-\bar{z}_1)dz}{z}-\frac{(z-z_1)(\bar{z}-\bar{z}_1)d\bar{z}}{\bar{z}}$ are closed and hence vanish upon application of Stokes' Theorem. In short, we recover the calculation given by the OP, but now for a pair of singularities.
I hope someone will improve on my non-answer here.
EDIT: (12-16-20), indeed, as per Steven Gubkin's comment, my "suspect" closed forms do not seem to be closed. I also calculate:
$$ d\beta = d \left[\frac{z\bar{z}dz}{z-z_1}-\frac{z\bar{z}d\bar{z}}{\bar{z}-\bar{z}_1} \right] = \left( \frac{\bar{z}}{\bar{z}-\bar{z}_1} + \frac{z}{z-z_1} \right) d\bar{z} \wedge dz$$
Then, I thought, maybe this integrates to zero for the area $|z| \leq R$. Well,
after some algebra,
$$ d\beta = \frac{4i(x^2+y^2+xx_1)}{(x-x_1)^2+(y-y_1)^2} \, dx \wedge dy. $$
If $z \neq z_1$ (as I have always implicitly assumed) then the denominator is positive over the disk $|z| \leq R$ and only the $xx_1$-term vanishes in the integral. It looks to me like that integral does not vanish. Apparently, I cannot have my cake and eat my antiholomorphic cake too.
