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?

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:
$$\gamma_5: \left[0, \frac{5\pi}{4}\right] \to \mathbb{R}^2,\ \gamma_5(t) = (\pi(1+\sqrt{2} \cos t), \pi \sqrt{2} \sin t)$$
$$\gamma_6: \left[\frac{3\pi}{2}, \pi(1+\sqrt{2})\right] \to \mathbb{R}^2,\ \gamma_6(t) = (t,0)$$
