hey mate,

Employ Euler's Identity

e^iz = cos(z) + isin(z)

thus,

e^-iz = cos(z) -isin(z)

thus,

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,

David