Updated greek laters @ latex compiling

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Let's write Newton's second law:
$$ \left\{\begin{matrix}
− \frac{du_x}{dt} = \frac{F \sin \varphi }{m}& \\
+ \frac{du_x}{dt} = \frac{F \sin \varphi }{m}& \\
\frac{du_y}{dt} = \frac{F (1- \cos \varphi) }{m} &
\end{matrix}\right. $$
$$ \left\{\begin{matrix}
− \frac{d}{dt} (u \sin \varphi)= \frac{F \sin \varphi }{m}& \\
+ \frac{d}{dt} (u \sin \varphi)= \frac{F \sin \varphi }{m}& \\
\frac{d}{dt} (u \cos \varphi) = \frac{F (1- \cos \varphi) }{m} &
\end{matrix}\right. $$
We solve the system of differential equations:
$$ \left\{\begin{matrix}
− \frac{d\varphi}{dt} = \frac{F \sin \varphi }{mu}& (a)\\
+ \frac{d\varphi}{dt} = \frac{F \sin \varphi }{mu}& (a)\\
\frac{du}{dt} = \frac{F (1- \cos \varphi) }{m} & (b)
\end{matrix}\right. $$
We divide $(b)$ by $(a)$:
−$$ \frac{du}{d\varphi} = u\frac{1- \cos \varphi}{\sin \varphi} $$
+$$ \frac{du}{d\varphi} = u\frac{1- \cos \varphi}{\sin \varphi} $$
−$$ \frac{du}{u} = \frac{1- \cos \varphi}{\sin \varphi} d\varphi $$
+$$ \frac{du}{u} = \frac{1- \cos \varphi}{\sin \varphi} d\varphi $$
$$ \frac{du}{u} = tg (\frac{\varphi}{2}) d\varphi $$
We integrate both parts of the equation:
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