How prove this integral $\int_0^{2\pi }{\frac{{{e^{\cos x}}\cos\left({\sin x}\right)}}{{p-\cos\left({y-x}\right)}}}dx$ show that 

$$ \int\limits_0^{2\pi }{\frac{{{e^{\cos x}}\cos\left({\sin x}\right)}}{{p-\cos\left({y-x}\right)}}}dx =\frac{{2\pi }}{{\sqrt{{p^2}-1}}}\exp\left({\frac{{\cos y}}{{p+\sqrt{{p^2}-1}}}}\right)\cos\left({\frac{{\sin y}}{{p+\sqrt{{p^2}-1}}}}\right);\left({p > 1}\right) $$

I think this is nice integral,But I can't show it,Thank you 
 A: Observe that the numerator is simply the real part of $e^{e^{i x}}$.  Thus, the desired integral is simply the real part of the contour integral
$$2 i \oint_{|z|=1} dz \frac{e^z}{e^{-i y} z^2-2 p z + e^{i y}}$$
(This is derived by substituting $z=e^{i x}$, $dx = -i dz/z$, $\cos{x}=(z+z^{-1})/2$, $\sin{x}=(z-z^{-1})/(2 i)$, and doing a little algebra.)
The poles of the integrand are at $z_{\pm}=(p \pm \sqrt{p^2-1}) e^{i y}$, of which only $z_-$ is inside the unit circle (recall that $p \gt 1$).  The residue at this pole is simply
$$2 i\frac{e^{(p-\sqrt{p^2-1}) e^{i y}}}{-2 \sqrt{p^2-1}}$$
and the integral is therefore, by the residue theorem, $i 2 \pi$ times this residue, or  
$$\frac{2 \pi}{\sqrt{p^2-1}} e^{(p-\sqrt{p^2-1}) e^{i y}}$$
We then take the real part of the above to get the sought-after integral.  Thus,
$$\int_0^{2 \pi} dx \frac{e^{\cos{x}} \cos{(\sin{x})}}{p-\cos{(y-x)}} = \frac{2 \pi}{\sqrt{p^2-1}} e^{\left (p-\sqrt{p^2-1}\right ) \cos{y}} \cos{\left [\left (p-\sqrt{p^2-1}\right ) \sin{y}\right ]} $$
as was to be shown.
