Solving what Mathematica could not Right, so as the final step of my project draws near and after having made a bad layout sort of question, I am posting a new one very clear and unambiguous. I need to find this specific definite integral which Mathematica could not solve:
$$  \int_{x=0}^\pi \frac{\cos \frac{x}{2}}{\sqrt{1 + A \sin^2 \frac{x}{2}}} \sin \left( \frac{1+A}{\sqrt{A}} \omega \tanh^{-1} \frac{\cos \frac{x}{2}}{\sqrt{1 + A \sin^2 \frac{x}{2}}} \right) \, dx.$$
where $ 0<A<1 $ and $ \omega > 0 $ are parameters of the problem.
I tried to use a substitution of the argument of the hyperbolic arctan but that seemed to make it worse. I was posting here in the hopes of receiving help, if someone could tell me if it is at all possible to solve it analytically via a trick of sorts or a clever substitution, or maybe it is an elliptic integral in disguise. I thank all helpers.
** My question on the Melnikov integral can be found here I just used trig identities to soften it up

Let $\mathcal{I}$ be the integral at hand and $B = \frac{1+A}{\sqrt{A}}\omega$. Introduce variables $y, t, \theta, z$ such that
$$y = \sin\frac{x}{2},\quad t = \tanh\theta = \frac{\cos\frac{x}{2}}{\sqrt{1+A\sin^2\frac{x}{2}}}\quad\text{ and }\quad z = e^\theta$$
Notice
$$t = \sqrt{\frac{1-y^2}{1+Ay^2}} \implies 
y = \sqrt{\frac{1-t^2}{1+At^2}}
\implies 
\frac{dy}{\sqrt{1+Ay^2}} = - \frac{t\sqrt{A+1}dt}{\sqrt{1-t^2}(1+At^2)}
$$
and $dt = (1-t^2)d\theta$, we have
$$\begin{align}
\mathcal{I} &= 2\int_0^1 \sin(B\theta)\frac{dy}{\sqrt{1+Ay^2}}
= 2\sqrt{A+1}\int_0^1\sin(B\theta)\frac{t\sqrt{1-t^2}}{1+At^2}\frac{dt}{1-t^2}\\
&= 2\sqrt{A+1}\int_0^\infty \frac{\sin(B\theta)\sinh\theta}{\cosh^2\theta + A\sinh^2\theta} d\theta
= -i\sqrt{A+1}\int_{-\infty}^\infty \frac{e^{iB\theta}\sinh\theta}{\cosh^2\theta + A\sin^2\theta} d\theta\\
&= -i2\sqrt{A+1}\int_0^\infty \frac{z^{iB}(z^2-1)}{(z^2+1)^2 + A(z^2-1)^2} dz
\end{align}
$$
Let $\phi = \tan^{-1}\sqrt{A}$, this can be simplified as
$$
\mathcal{I} = 
-\frac{2i}{\sqrt{A+1}}\int_0^\infty \frac{z^{iB}(z^2-1)}{z^4 + 2\left(\frac{1-A}{1+A}\right) z^2 + 1} dz
= -2i\cos\phi\int_0^\infty \frac{z^{iB}(z^2-1)}{(z^2+e^{2i\phi})(z^2+e^{-2i\phi})} dz$$
Consider following contour integral
$$\mathcal{J(\epsilon,R)} \stackrel{def}{=} \oint_{C(\epsilon,R)} \frac{(-z)^{iB}(z^2-1)}{(z^2+e^{2i\phi})(z^2+e^{-2i\phi})}
dz
\tag{*1}$$
where 


*

*$\arg(-z) = 0$ on negative real axis.

*$(-z)^{iB}$ has a branch cut along positive real axis.

*$C(\epsilon,R)$ is the contour consists of 


*

*$C_1$ : line segment from $\epsilon \to R$ above the positive real axis.

*$C_2$ : circular arc $Re^{iu}$ for $u$ from $0 \to 2\pi$.

*$C_3$ : line segment $R \to \epsilon$ below the positive real axis.

*$C_4$ : circular arc $\epsilon e^{iu}$ for $u$ from $2\pi \to 0$.



It is easy to see in $\mathcal{J}(\epsilon,R)$, 


*

*the contribution from $C_1$ and $C_3$ adds up to
$$(e^{\pi B} - e^{-\pi B})\int_\epsilon^R \frac{z^{iB}(z^2-1)}{(z^2+e^{2i\phi})(z^2+e^{-2i\phi})} dz$$

*the contribution from $C_4$ vanishes as $\epsilon \to 0$.

*Since $B > 0$ is a real number, $|(-z)^{iB}|$ is bounded from above by $e^{\pi B}$ and the contribution from $C_2$ behaves as $O(R^{-1})$ as $R \to \infty$.


Combine these, we find
$$\mathcal{I} = -i\frac{\cos\phi}{\sinh(\pi B)}\lim_{\epsilon \to 0,R \to \infty} \mathcal{J}(\epsilon,R)$$
The integrand in $(*1)$ has 4 poles inside the contour: $\; e^{i(\frac{\pi}{2} \pm \phi)}\;$ and $\;e^{i(\frac{3\pi}{2} \pm \phi)}$.


*

*The residues at $e^{i(\frac{\pi}{2} \pm \phi)}$ are
$\displaystyle\;(e^{-(-\frac{\pi}{2} \pm \phi)B})\frac{-e^{\pm 2i\phi} - 1}{4i e^{\pm i\phi }(-e^{\pm 2i \phi} + \cos(2\phi))} = \mp\frac{e^{(\frac{\pi}{2}\mp \phi)B}}{4\sin\phi}$

*The residues at $e^{i(\frac{3\pi}{2} \pm \phi)}$ are
$\displaystyle\;(e^{-(\frac{\pi}{2} \pm \phi)B})\frac{-e^{\pm 2i\phi} - 1}{-4i e^{\pm i\phi }(-e^{\pm 2i \phi} + \cos(2\phi))} = \pm\frac{e^{(-\frac{\pi}{2}\mp \phi)B}}{4\sin\phi}$
This implies
$$\begin{align}
\mathcal{I} &= \left(-i \frac{\cos\phi}{\sinh\pi B}\right)\left(\frac{2\pi i}{4\sin\phi}\right)\left[
-e^{(\frac{\pi}{2}-\phi)B}
+e^{(\frac{\pi}{2}+\phi)B}
+e^{(-\frac{\pi}{2}-\phi)B}
-e^{(-\frac{\pi}{2}+\phi)B}
\right]\\
&= \left(\frac{\cos\phi}{\sinh\pi B}\right)\left(\frac{2\pi}{4\sin\phi}\right)
   (e^{\frac{\pi}{2}B} - e^{-\frac{\pi}{2}B})
   (e^{\phi B} - e^{-\phi B})
 = \frac{\pi}{\tan\phi}\frac{\sinh(B\phi)}{\cosh\left(\frac{\pi}{2}B\right)}\\
&= \frac{\pi}{\sqrt{A}}\frac{\sinh(B\tan^{-1}\sqrt{A})}{\cosh\left(\frac{\pi}{2}B\right)}
\end{align}
$$
Treating $A, B$ as two independent parameters, in the limiting case $A \to 0$, we find
$$\lim_{A\to 0} \mathcal{I} = \frac{\pi B}{\cosh\left(\frac{\pi}{2}B\right)}$$
This matches what first pointed out by @nospoon  in the comments.
