easily be integrated: $${\frac{\raisebox{-2pt}{$s^2$}}{1 - s^2}} = {\frac{\raisebox{-2pt}{$1 - c^2$}}{c^2}} = f^2(\chi)\, {\mathrm{e}}^{(2{\epsilon}- 1)
  \phi}\ .$$ The positive integration constant $f^2$ may depend on $\chi$. These expressions we plug into the sum of and , $$\begin{aligned}
  0 & = {\mathrm{e}}^{\phi/2} {\partial}_\chi c^2 + {\partial}_{\phi} (cs) - {\tfrac{1}{2}}(2{\epsilon}- 1) cs
    \notag \\[2pt]
  & = - {\frac{\raisebox{-2pt}{$2 f\, {\mathrm{e}}^{(3 - 4{\epsilon})\phi/2}$}}{(f^2 + {\mathrm{e}}^{(1 - 2{\epsilon})\phi}
    )^2}}\ \big[ {\partial}_\chi f \pm {\tfrac{1}{2}}(2{\epsilon}- 1)\, {\mathrm{e}}^{({\epsilon}- 1)\phi}
    f^2 \big]\ ,
 \end{aligned}$$ where the sign ambiguity originates from taking the square root of $(cs)^2$. The equation is satisfied if the expression in square brackets vanishes. For ${\epsilon}=-1$, this is only possible if $f=0$ since $f$ is independent of $\phi$. For ${\epsilon}=+1$, which corresponds to $O\in
\mathrm{SO}(2)$, we find $$f = 0 \quad\text{or}\quad f = \pm {\frac{\raisebox{-2pt}{$2$}}{\chi - \chi_0}}\ ,$$ with $\chi_0$ an integration constant. $f=0$ implies $c=\pm 1$ and $s=
0$. For nontrivial $f$ we obtain (with the relative sign fixed by the original equations and $\eqref{O2_2}$) $$c = \pm \frac{\chi - \chi_0}{\sqrt{4 {\mathrm{e}}^\phi + (\chi - \chi_0)^2\,}
  \,}\ ,\quad s = \pm {\frac{\raisebox{-2pt}{$2\, {\mathrm{e}}^{\phi/2}$}}{\sqrt{4 {\mathrm{e}}^\phi + (\chi -
  \chi_0)^2\,}\,}}\ ,$$ or in terms of $\tau'=(\chi-\chi_0)+2{\mathrm{i}}\,{\mathrm{e}}^{\phi/2}$, $$c + {\mathrm{i}}s = \pm {\frac{\raisebox{-2pt}{$\tau'$}}{|\tau'|}}\ .$$

[99]{} J. Bagger 