I am interested in various solutions to this integral, here is one of the versions:
$$...=\text{Re}\int\limits_0^{2\pi}e^{-\sin^2x+i\left(6x-\frac{\sin(2x)}{2}\right)}dx=e^{-\frac{1}{2}}\text{Re}\int\limits_0^{2\pi}e^{\frac{\cos(2x)}{2}+6ix-\frac{i\sin(2x)}{2}}dx=$$$$=e^{-\frac{1}{2}}\text{Re}\int\limits_0^{2\pi}e^{6ix+\frac{e^{-2ix}}{2}}dx=e^{-\frac{1}{2}}\text{Re}\int\limits_{|z|=1}\frac{z^6\cdot e^{\frac{1}{2z^2}}}{iz}dz=-e^{-\frac{1}{2}}\text{Re}\ i\int\limits_{|z|=1}z^5\cdot e^{\frac{1}{2z^2}}dz=$$$$=2\pi e^{-\frac{1}{2}}\text{Re}\left(\mathrm{Res}_0z^5\cdot e^{\frac{1}{2z^2}}\right)=\frac{2\pi e^{-\frac{1}{2}}}{48}=\frac{\pi}{24\sqrt e}.$$