With $x > y > 0$, I am looking for an analytical expression for $$\int_{0}^{\pi} \! \frac{\sin \sqrt{x + y \cos\phi}}{\sqrt{x + y \cos\phi}} \, d\phi$$ which I suspect may be elementary, or Bessel-like. I did not find the integral in Gradshteyn & Ryzhik or similar.
Expanding $\frac{\sin z}{z}$ yields the series $$\int_{0}^{\pi} \! \frac{\sin \sqrt{x + y \cos\phi}}{\sqrt{x + y \cos\phi}} \, d\phi = \sum_{k=0}^{\infty} \frac{\Gamma(\tfrac{1}{2}) \, \Gamma(k+\tfrac{1}{2})}{k! \, (4k+1)!} \, _1F_2\bigl(1; 2k+1, 2k+\tfrac{3}{2}; -\tfrac{x}{4}\bigr) \, y^{2k}$$ which I cannot evaluate further. There are other infinite series that I found similarly intractable.
I thought the residue theorem might be a good approach, $$\int_{0}^{\pi} \! \frac{\sin \sqrt{x + y \cos\phi}}{\sqrt{x + y \cos\phi}} \, d\phi = \oint\limits_{|z|=1} \! \frac{\sin \sqrt{x + y \, (z+1/z)/2}}{\sqrt{x + y \, (z+1/z)/2}} \, \frac{dz}{2iz}$$ with poles inside the contour at $z = 0$ and $z = \frac{\sqrt{x^2-y^2} - x}{y}$, but I cannot get the residues except again in infinite series form.
At the risk of making matters less transparent, here is why I was expecting perhaps a Bessel-like special function to pop out.
Let $x = a^2 + b^2 - 2 a b \cos\alpha \cos\beta$, $y = 2 a b \sin\alpha \sin\beta$. Then we can use the addition theorem for spherical Bessel functions, $$\frac{\sin \sqrt{x + y \cos\phi}}{\sqrt{x + y \cos\phi}} = \sum_{l=0}^{\infty} (2l+1) \, j_l(a) \, j_l(b) \, P_l(\cos\alpha \cos\beta + \sin\alpha \sin\beta \cos\phi)$$ with $P_l$ the Legendre polynomial, and integrate out $\phi$ to find $$\frac{1}{\pi} \int_{0}^{\pi} \! \frac{\sin \sqrt{x + y \cos\phi}}{\sqrt{x + y \cos\phi}} \, d\phi = \sum_{l=0}^{\infty} (2l+1) \, j_l(a) \, j_l(b) \, P_l(\cos\alpha) \, P_l(\cos\beta) \;.$$ So this is like a two-dimensional variant of the spherical Bessel function $j_0$.