Is it possible to evaluate analytically the following nontrivial triple integral? In a physical mathematical problem, I came across a nontrivial triple integral below obtained upon 3D inverse Fourier transformation. It would be great if someone here could provide with some hints that could help to evaluate analytically
$$
I =\frac{1}{(2\pi)^3} \int_0^{2\pi} \int_0^\infty \int_0^{\pi}
  \frac{\sin\theta \sin^2\phi}{a \cos^2 \theta + b \sin^2 \theta} \, e^{ikh\sin\theta\cos\phi}  \, \mathrm{d} \theta \, \mathrm{d} k \, \mathrm{d} \phi \, .
$$
where $a$ and $b$ are two positive real numbers.
Using the change of variable $q=\cos\theta$, the latter equation can be written as
$$
I = \frac{1}{(2\pi)^3} \int_0^{2\pi} \int_0^\infty \int_{-1}^{1} 
  \frac{\sin^2\phi}{(a-b)q^2+b} \, e^{ikh\sqrt{1-q^2}\cos\phi} \,  \mathrm{d} q \, \mathrm{d} k \, \mathrm{d} \phi \, .
$$
A guess solution using Maple for some numerical values for $a$ and $b$ is obtained as
$$
I = \frac{1}{4\pi h \sqrt{ab}} \, .
$$
But, is there a way to prove that really? Any idea / feedback is welcome. 
Thanks a lot!
Fede
 A: Well, for the first integral we have:
$$\mathscr{I}:=\int_0^{2\pi}\int_0^\infty\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)\cdot\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\theta\space\text{d}\text{k}\space\text{d}\phi\tag1$$
Using Fubini's theorem:
$$\mathscr{I}:=\int_0^{2\pi}\int_0^\pi\int_0^\infty\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)\cdot\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\text{k}\space\text{d}\theta\space\text{d}\phi=$$
$$\int_0^{2\pi}\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\int_0^\infty\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)\space\text{d}\text{k}\space\text{d}\theta\space\text{d}\phi\tag2$$
Now, we can use:
$$\int_0^\infty\exp\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{k}\cdot\text{h}\cdot i\right)\space\text{d}\text{k}=\frac{i}{\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{h}}\tag3$$
when $\Im\left(\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{h}\right)>0$
So, we get:
$$\mathscr{I}=\int_0^{2\pi}\int_0^\pi\frac{\sin\left(\theta\right)\cdot\sin^2\left(\phi\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\cdot\frac{i}{\sin\left(\theta\right)\cdot\cos\left(\phi\right)\cdot\text{h}}\space\text{d}\theta\space\text{d}\phi=$$
$$\int_0^{2\pi}\frac{\sin^2\left(\phi\right)\cdot i}{\cos\left(\phi\right)\cdot\text{h}}\int_0^\pi\frac{\sin\left(\theta\right)}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\cdot\frac{1}{\sin\left(\theta\right)}\space\text{d}\theta\space\text{d}\phi=$$
$$\int_0^{2\pi}\frac{\sin^2\left(\phi\right)\cdot i}{\cos\left(\phi\right)\cdot\text{h}}\int_0^\pi\frac{1}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\theta\space\text{d}\phi\tag4$$
Now, use:
$$\int_0^\pi\frac{1}{\text{a}\cdot\cos^2\left(\theta\right)+\text{b}\cdot\sin^2\left(\theta\right)}\space\text{d}\theta=\frac{\pi}{\sqrt{\text{a}\cdot\text{b}}}\tag5$$
So, we get:
$$\mathscr{I}=\int_0^{2\pi}\frac{\sin^2\left(\phi\right)\cdot i}{\cos\left(\phi\right)\cdot\text{h}}\cdot\frac{\pi}{\sqrt{\text{a}\cdot\text{b}}}\space\text{d}\phi=\frac{\pi\cdot i}{\text{h}\cdot\sqrt{\text{a}\cdot\text{b}}}\int_0^{2\pi}\frac{\sin^2\left(\phi\right)}{\cos\left(\phi\right)}\space\text{d}\phi\tag6$$
