Finding a specific improper integral on a solution path to a 2 dimensional system of ODEs In my study of dynamical systems I was recently met with this system of ODEs:

$ \dot{x}=\frac{\sinh{(y)}}{\cosh{(y)}+A\cos{(x)}} $
$ \dot{y}=\frac{A\sin{(x)}}{\cosh{(y)}+A\cos{(x)}} $

for a parameter A satisfying $ 0 < A < 1 $
Now clearly $ (0,0) $ and $ (2\pi,0) $ are fixed points and we may check there is a solution (heteroclinic) path (curve) from $ (0,0) $ to $ (2\pi,0) $. I need to find its representation not as a curve but via time dependency meaning (x(t),y(t)), I also know via simple integration that the solution sits on the equation curve $ \cosh{(y)} = -A\cos{(x)} + (A+1) $ but I cannot solve these equations for x,y components specifically as it seems impossible. Finally I need to determine if the following Melnikov integral has simple zeros: $ M(t_0)=\int_{-\infty}^{\infty} -\frac{A}{A+1}\sin{(x(t))}\sin{(\omega(t+t_0))} dt $, where x(t) is the x component of the solution curve mentioned earlier. I do not have $ x(t) $ explicitly let alone handle this integral. Maybe it can be handled via perturbation or numerical analysis (I only need to know if it has any simple zeros). I figured a good first step was to solve for x(t) via the equation $ \dot{x}=\frac{\sinh{(y)}}{\cosh{(y)}+A\cos{(x)}} = \frac{\sqrt{(A\cos{(x)}-A-1)^2-1}}{A+1} $ which is the equation on the solution curve.
 A: HINT :
Equation of the trajectory : $y(x)=\cosh^{-1}\left(-A\cos(x)+A+1\right)$
$$y(x)=\ln \left|\left(-A\cos(x)+A+1\right)\pm\sqrt{\left(-A\cos(x)+A+1\right)^2-1}\:\right|$$
Implicit equation of the motion : 
$$t=(A+1)\int \frac{dx}{\sqrt{\left(-A\cos(x)+A+1\right)^2-1}} $$
After some calculus with help of WolframAlpha :
$$t=-(A+1)\sqrt{\frac{2}{A(1-\cos(x))}}\sin(\frac{x}{2})\tanh^{-1}\left(\sqrt{\frac{2}{-A\cos(x)+A+2}}\cos(\frac{x}{2}) \right)+C$$
$C=$constant. For example, if $x=\pi$ at $t=t_m$ then $C=t_m$
As far as I know there is no closed form for the inverse function $x(t)$.
IN ADDITION :
$$ M(t_0)=\int_{-\infty}^{\infty} -\frac{A}{A+1}\sin{(x(t))}\sin{(\omega(t+t_0))} dt $$
$dt=(A+1)\frac{dx}{\sqrt{\left(-A\cos(x)+A+1\right)^2-1}} $
$$ M(t_0)=-A\int_0^{2\pi} \sin{(x)}\sin{(\omega(t(x)+t_0))} \frac{dx}{\sqrt{\left(-A\cos(x)+A+1\right)^2-1}}  $$
where $t(x)=-(A+1)\sqrt{\frac{2}{A(1-\cos(x))}}\sin(\frac{x}{2})\tanh^{-1}\left(\sqrt{\frac{2}{-A\cos(x)+A+2}}\cos(\frac{x}{2}) \right)$
