Solving what Mathematica could not
Solution 1:
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.
Solution 2:
This is not a full answer.From answer of user You're In My Eye .First, let's make the following substitutions:
$$y=\sin \frac{x}{2}$$
$$B=\frac{1+A}{\sqrt{A}} \omega$$
$$2 \int_0^1 \sin \left( B~ \text{arctanh}~\sqrt{\frac{1-y^2}{1+Ay^2}} \right) \frac{dy}{\sqrt{1+Ay^2}}$$
substitutions:
$$y={\frac { \sqrt{- \left( A{t}^{2}+1 \right) \left( {t}^{2}-1
\right) }}{A{t}^{2}+1}}
$$
Then we obtain:
$$2\, \sqrt{A+1}\int_{0}^{1}\!{\frac {\sin \left( B{\rm arctanh} \left(t
\right) \right) t}{ \left( A{t}^{2}+1 \right) \sqrt{-{t}^{2}+1}}}
\,{\rm d}t
$$
substitutions:
$$t=\tanh \left( k \right) $$
have :
$$2\,\sqrt {A+1}\int_{0}^{\infty }\!{\frac {\sin \left( Bk \right) \sinh
\left( k \right) }{A \left( \cosh \left( k \right) \right) ^{2}+
\left( \cosh \left( k \right) \right) ^{2}-A}}\,{\rm d}k
$$
trig identity:
cosh(k)^2-sinh(k)^2 = 1
and A+1=m
$$2\, \sqrt{A+1}\int_{0}^{\infty }\!{\frac {\sin \left( Bk \right) \sinh
\left( k \right) }{1+ \left( A+1 \right) \left( \sinh \left( k
\right) \right) ^{2}}}\,{\rm d}k \tag{1}
$$
I have a simple form of integral:
$$2\, \sqrt{m}\int_{0}^{\infty }\!{\frac {\sin \left( B k \right) \sinh
\left( k \right) }{1+m \left( \sinh \left( k \right) \right) ^{2}}}
\,{\rm d}k
$$
Substitutions back $$B=\frac{1+A}{\sqrt{A}} \omega$$ to equation 1.
I have:
$$2\, \sqrt{A+1}\int_{0}^{\infty }\!{\frac {\sin \left(\frac{1+A}{\sqrt{A}} \omega k \right) \sinh
\left( k \right) }{1+ \left( A+1 \right) \left( \sinh \left( k
\right) \right) ^{2}}}\,{\rm d}k
$$
Mathematica can find solution for this integral.
A = 1/4;
omega = 1;
int = Normal[2*Sqrt[A + 1]*Integrate[(Sin[(1 + A)/Sqrt[A]*omega*k]*Sinh[k])/(
1 + (A + 1)*Sinh[k]^2), {k, 0, Infinity}]]
(*(1/1769)2 Sqrt[5] ((305 -
122 I) ((1/5 + (3 I)/20) Hypergeometric2F1[1, 1 - (5 I)/2,
2 - (5 I)/2, -((1 + 2 I)/Sqrt[5])] + (1/5 - (3 I)/
20) Hypergeometric2F1[1, 1 - (5 I)/2,
2 - (5 I)/2, -((1 - 2 I)/Sqrt[5])] + (1/5 - (3 I)/
20) Hypergeometric2F1[1, 1 - (5 I)/2, 2 - (5 I)/2, (1 - 2 I)/
Sqrt[5]] + (1/5 + (3 I)/20) Hypergeometric2F1[1, 1 - (5 I)/2,
2 - (5 I)/2, (1 + 2 I)/Sqrt[5]]) + (5 +
2 I) (61 ((1/5 + (3 I)/20) Hypergeometric2F1[1, 1 + (5 I)/2,
2 + (5 I)/2, -((1 + 2 I)/Sqrt[5])] + (1/5 - (3 I)/
20) Hypergeometric2F1[1, 1 + (5 I)/2,
2 + (5 I)/2, -((1 - 2 I)/Sqrt[5])] + (1/5 - (3 I)/
20) Hypergeometric2F1[1, 1 + (5 I)/2, 2 + (5 I)/2, (
1 - 2 I)/Sqrt[5]] + (1/5 + (3 I)/20) Hypergeometric2F1[1,
1 + (5 I)/2, 2 + (5 I)/2, (1 + 2 I)/Sqrt[5]]) + (2 +
5 I) ((6 +
5 I) ((1/5 + (3 I)/20) Hypergeometric2F1[1, 3 - (5 I)/2,
4 - (5 I)/2, -((1 + 2 I)/Sqrt[5])] + (1/5 - (3 I)/
20) Hypergeometric2F1[1, 3 - (5 I)/2,
4 - (5 I)/2, -((1 - 2 I)/Sqrt[5])] + (1/5 - (3 I)/
20) Hypergeometric2F1[1, 3 - (5 I)/2, 4 - (5 I)/2, (
1 - 2 I)/Sqrt[5]] + (1/5 + (3 I)/20) Hypergeometric2F1[
1, 3 - (5 I)/2, 4 - (5 I)/2, (1 + 2 I)/Sqrt[5]]) - (6 -
5 I) ((1/5 + (3 I)/20) Hypergeometric2F1[1, 3 + (5 I)/2,
4 + (5 I)/2, -((1 + 2 I)/Sqrt[5])] + (1/5 - (3 I)/
20) Hypergeometric2F1[1, 3 + (5 I)/2,
4 + (5 I)/2, -((1 - 2 I)/Sqrt[5])] + (1/5 - (3 I)/
20) Hypergeometric2F1[1, 3 + (5 I)/2, 4 + (5 I)/2, (
1 - 2 I)/Sqrt[5]] + (1/5 + (3 I)/20) Hypergeometric2F1[
1, 3 + (5 I)/2, 4 + (5 I)/2, (1 + 2 I)/Sqrt[5]]))))*)
Solution 3:
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 $$\tanh^{-1} \frac{\cos \frac{x}{2}}{\sqrt{1+ a \sin^2 \frac{x}{2}}} \mapsto x. \tag{5}$$
Solution 4:
I highly doubt this integral has any useful analytic form. I suggest you either use numerical integration or simplify your initial model to obtain a solvable integral.
Using @tired 's suggestion, we can make you integral much easier. First, let's make the following substitutions:
$$y=\sin \frac{x}{2}$$
$$B=\frac{1+A}{\sqrt{A}} \omega$$
Then we obtain:
$$I(A,B)=2 \int_0^1 \sin \left( B~ \text{arctanh}~\sqrt{\frac{1-y^2}{1+Ay^2}} \right) \frac{dy}{\sqrt{1+Ay^2}}$$
For the most simple case Mathematica gives:
$$B=1,~~A=0$$
$$I(0,1)=2\pi \frac{\sinh (\pi/2)}{\sinh \pi}$$
Mathematica can't take this integral even with $B=1,~A=1$. Neither with $A=0$ and $B$ not defined.
Which is why I think numerical methods is the only way for you. Numerically, the integral is very nice (and very weakly depends on $A$ by the way).
See the plots below for $I(A,B)$, for $A \in (0,1)$ and $B \in (0,4)$:
I'm aware this is not even close to answering your question, but you might find this useful.
And yes, there might be some connection with elliptic integrals, as you can see by the argument of the inverse hyperbolic tangent.
Edit
I was wrong - this integral is workable. See @MariuszIwaniuk 's much better answer!
And see @achillehui's answer for the final solution. With permission, I add the 3D plot of the exact solution to compare with numerical results:
All credit goes to achille hui!