Employ Euler's Identity
e^iz = cos(z) + isin(z)
e^-iz = cos(z) -isin(z)
sin(z) = (1/2i)*(e^iz + e^-iz))
hence your integral becomes
(1/2i)* (e^iz + e^-iz)/(z^2 + 6z + 10) = (1/2i)( e^iz/(z^2 + 6z + 10) + e^-iz/(z^2 + 6z + 10))
obviously you have two integrals here, which unfortunately in split into 4 integrals as 1/(z^2 + 6z + 10) needs to factorised into the form 1/(z-a)(z-b) and then resolved into partial fractions, when this is achieved the Cauchy Residue Thm can be applied to evaluate the integral.
Hope this helps,