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

Here is a sketch for a residue-free solution: 
First, consider the formula $$\int_0^{\infty} \frac{\cos( a x)}{\cosh(\frac{\pi}{2} x)}dx = \operatorname{sech} a. \tag{1}$$
Since $\displaystyle \,\,\sin(a)\sin(b) = \frac12 \cos(a-b)-\frac12 \cos(a+b), \,\,$ we obtain 
$$ \int_0^{\infty} \frac{\sin( a x) \sin(b x)}{\cosh(\frac{\pi}{2} x)}dx = \frac12 \operatorname{sech}(a-b) - \frac12 \operatorname{sech}(a+b). \tag{2}$$
Now, using a fourier inversion argument $\displaystyle\left( f(a)=\int_0^{\infty} g(x)\sin(a x) dx \iff \frac{\pi}{2} g(a) = \int_0^{\infty} f(x) \sin(a x) dx \right),\,$ we obtain 
 $$\int_0^{\infty} \sin( b x) ( \operatorname{sech}(a-x) - \operatorname{sech}(a+x)) dx = \pi \cfrac{ \sin(a b)}{\cosh( \frac{\pi}{2} b)}. \tag{3}$$
Finally, letting $a \mapsto i \tan^{-1} \sqrt{a},\,$ and noting that 
$$\operatorname{sech}(a-b) - \operatorname{sech}(a+b) = 2 \frac{\sinh a}{\cosh^2 a} \,\frac{\sinh b}{\cosh^2 b}\, \frac1{1-\tanh^2 a \, \tanh^2 b},$$
we find that 
$$\frac{\pi}{\sqrt{a}} \cfrac{\sinh( b \tan^{-1} \sqrt{a} )}{\cosh( \frac{\pi}{2} b)} = 2 \sqrt{1+a} \int_0^{\infty} \cfrac{\sin(b x) \sinh x}{\cosh^2 x + a \, \sinh^2 x} dx, \tag{4}$$
which is exactly your integral, with $b= \frac{1+a}{\sqrt{a}} \omega,$ and the substitution 
$$\tanh^{-1} \frac{\cos \frac{x}{2}}{\sqrt{1+ a \sin^2 \frac{x}{2}}} \mapsto x. \tag{5}$$
