Green's theorem and translated angular form on the path $\gamma:[0,3\pi/2]\to\Bbb R^2,\gamma(t)=(t,\pi\cos t)$-strange result 
I would like to compute the following integral $$\int_\gamma-\frac{y}{(x-\pi)^2+y^2}dx+\frac{x-\pi}{(x-\pi)^2+y^2}dy,$$ where $\gamma:\left[0,\frac{3\pi}2\right]\to\Bbb R^2,\gamma(t)=(t,\pi\cos t).$

Here is my answer which I would like to verify.
The differential $1$-form $\omega=-\frac{y}{(x-\pi)^2+y^2}dx+\frac{x-\pi}{(x-\pi)^2+y^2}dy$ is a translated angular form centered at $(\pi,0),$ hence $d\omega=0,$ that is $\omega$ is closed. I would like to apply Green's theorem, so, for that purpose, I considered the following paths:
$$\gamma_2:[0,1]\to\Bbb R^2,\gamma_2(t)=\left(\frac{3\pi}2,t\right),\\\gamma_3:[0,1]\to\Bbb R^2,\gamma_3(t)=\left(\frac{3\pi}2-\frac{3\pi}2t,\pi\right),\\\gamma_4:[0,2\pi]\to\Bbb R^2,\gamma_4(t)=(\pi+\cos t,-\sin t).$$ First, $$\int_{\gamma_2}\omega=\int_0^1\frac{\frac{3\pi}2-\pi}{\left(\frac{3\pi}2-\pi\right)^2+t^2}dt=\arctan\frac{t}{\frac{\pi}2}\Big|_0^1=\arctan\frac2\pi\\\int_{\gamma_3}\omega=\int_0^1\frac{\pi}{(t-\pi)^2+\pi^2}dt=\arctan\frac{t-\pi}\pi\Big|_0^1=\arctan\frac{1-\pi}\pi+\frac\pi4$$ and $$\int_{\gamma_4}\omega=-2\pi$$ because $\gamma_4$ is closed and negatively oriented. Now, let $D$ be the set "inside" the contour $\gamma+\gamma_2+\gamma_3$ and outside the circle $\gamma_4.$ By Green's theorem, $$\begin{aligned}\int_\gamma\omega+\int_{\gamma_2+\gamma_3+\gamma_4}\omega&=\int_{\partial D}\omega=\int_D d\omega=0\\\implies \int_\gamma\omega&=-\int_{\gamma_2+\gamma_3+\gamma_4}\omega=2\pi-\arctan\frac2\pi-\arctan\frac{1-\pi}\pi-\frac\pi4\\&=\frac{7\pi}4-\arctan\left(\tan\left(\arctan\frac2\pi+\arctan\frac{1-\pi}\pi\right)\right)\\&=\frac{7\pi}4-\arctan\left(\frac{\frac2\pi+\frac{1-\pi}\pi}{1-\frac2\pi\cdot\frac{1-\pi}\pi}\right)\\&=\frac{7\pi}4+\arctan\frac{(\pi-3)\pi}{\pi^2+2\pi-2}\\&=\arctan\left(\tan\left(\frac{7\pi}4+\arctan\frac{(\pi-3)\pi}{\pi^2+2\pi-2}\right)\right)\\&=\arctan\left(\frac{-1+\frac{(\pi-3)\pi}{\pi^2+2\pi-2}}{1+\frac{(\pi-3)\pi}{\pi^2+2\pi-2}}\right)\\&=\arctan\frac{2-5\pi}{2\pi^2-\pi-2}\end{aligned}$$
But I'm unsure as I didn't expect such a result. Did I make any mistakes?
 A: Your $\gamma_2$ ends at $\left(\frac{3\pi}{2},1\right)$, but presumably you want it to end at $(\frac{3\pi}{2},\pi)$ to connect to the start of $\gamma_3$. Correction:
$$ \gamma_2:[0,\pi]\to\mathbb{R}^2,\ \gamma_2(t)=\left(\frac{3\pi}2, t\right) $$
$$ \int_{\gamma_2} \omega = \int_0^\pi \frac{\frac{3\pi}2-\pi}{\left(\frac{3\pi}2-\pi\right)^2+t^2}\, dt = \arctan\frac{t}{\frac{\pi}{2}} \Big|_0^\pi = \arctan 2 $$
(Or you could have $\gamma_2$ keep domain $[0,1]$ and use $y=\pi t$, but then don't forget $dy = \pi\, dt$.)
You didn't properly substitute $x = \frac{3\pi}{2} - \frac{3\pi}{2} t$ when integrating on $\gamma_3$. It's also missing the factor from $dx = -\frac{3\pi}{2} dt$ and/or the minus sign at the beginning of the $dx$ term in the $\omega$ definition. Correction:
$$ \begin{align*} \int_{\gamma_3} \omega &= \int_0^1 \frac{-\pi}{\left(\frac{3\pi}{2} - \frac{3\pi}{2} t - \pi\right)^2 + \pi^2}\left(-\frac{3\pi}{2}\, dt\right)
 = \int_0^1 \frac{\frac{3}{2}}{\left(\frac{1}{2} - \frac{3}{2} t\right)^2 + 1}\, dt \\
\int_{\gamma_3} \omega &= \left. \arctan\left(\frac{3}{2} t - \frac{1}{2}\right) \right|_0^1 = \arctan 1 + \arctan \frac{1}{2}
\end{align*} $$
You are correct about the combination of the curves for Green's theorem:
$$ \int_\gamma \omega = - \int_{\gamma_2+\gamma_3+\gamma_4} \omega = 2\pi - \arctan 2 - \arctan 1 - \arctan \frac{1}{2} $$
Note $\arctan 2 + \arctan \frac{1}{2} = \frac{\pi}{2}$ since they form two angles of a right triangle.
$$ \int_\gamma \omega = \frac{5 \pi}{4} $$
And this makes sense since it equals the counter-clockwise change of angle from $(0,\pi)$ to $(\frac{3\pi}{2}, 0)$ with respect to the "center" point $(\pi, 0)$ of the translated angular form $\omega$.
Also, here are different paths I might have used, which are even easier to integrate:
$$\gamma_5: \left[0, \frac{5\pi}{4}\right] \to \mathbb{R}^2,\ \gamma_5(t) = (\pi(1+\sqrt{2} \cos t), \pi \sqrt{2} \sin t)$$
$$\gamma_6: \left[\frac{3\pi}{2}, \pi(1+\sqrt{2})\right] \to \mathbb{R}^2,\ \gamma_6(t) = (t,0)$$
A: Definitions are found in Multilinear_form.
I hope You can make it through this longer definitions page set.
There are of course several definitions for 1-form on the page!
The task is if You did not miss it for the last section: Integration of differential forms and Stokes' theorem for chains.
The $A=\mathbb R^2$. The border is $\gamma$.
From the first definition, this is

From Your second definition for $\gamma$, this is an ellipsis around the origin!
The other borders are a straight line through a point, $\gamma_{3}$ is a circle around the point $(\pi,0)$.
Form the given information there is a function
$f(x,y)=\arctan(\frac{x-\pi}{y})$
So to match the Stoke's theorem we need the pair $D$ and $\partial D$. The latter should be the boundary of $D$. And the boundary means unique only a sense of with way round may remain.
So the given straight lines have no $D$. They are no chains for example around a square. The circle and the ellipsis are chains closed around a region.
On the page Stokes theorem the connection to the Green's theorem is made. And there is given how to write a chain.
The first formula is the $d\omega$. The $f(x,y)$ of me is the $\omega$. That should be clear from the definitions.
For the
$\int_{\partial C}\omega$
The path $\partial C$ has to be parametrized.
$x(t)=\sin(t)$
$y(t)=\cos(t)$
parametrizes a circle around the origin starting at the point $(0,1)$.
Have a look at line integral. Use the formula for the chain, and path in a scaler field.
If the path is around the origin and closed like for the circle around the origin we have with the main sentence of the integral and differential calculus that the value is zero.
For a path around a point different from zero the situation is different. This question is similar: integration on the circle.
In the real plane, the path is the square root of the t derivatives of the path parametric functions squared. In the case of the unit circle, this is one, and complicated to write in LaTex.
We have:
$f(x,y)=f(t)=\arctan(\frac{x-\pi}{y})=\arctan(\frac{sin(t)-\pi}{cos(t)})$
This is to be integrated from $0$ to $2\pi$.
The solution takes a page if expressed in terms of $\pi$.
The value is $-1.33236$.
That is an example of how to do it.
There is another misconception in Your question the path is given incorrectly:
I suppose it should look this way:
$\gamma_{3}:[0,\pi]\rightarrow \mathbb R^{2}$, $\gamma_{3}(t)=\frac{3\pi}{2}(1-t)$
The chain or path looks then:

For the 1-form the path has to be taken too. This is explained too in Line integral. But there is no path length like in the $\omega$-integral.
I think that is enough help. There is some little work left. Enjoy or perhaps make clear before the time runs out. Chain rule is common from school.
