Fourier series coefficient for sin(wt+theta)
Just pointers:


pull the \$\frac1{2j}\$ out of your integral, it's just a constant.
The integral is a linear operation – use that to split your sum, so you have two integrals; one over \$e^{j\omega t + j\theta}e^{-jk\omega t}\$ and one over \$e^{-j\omega t - j\theta}e^{-jk\omega t}\$.
\$e^{j\omega t + \theta}=e^{j\omega t}\cdot e^{j\theta}\$; this applies to both integral. \$e^{\pm j\theta}\$  is a constant and can be pulled out of your integral. 
You end up with an integrated product: \$e^{\pm j\omega t}\cdot e^{-jk\omega t}\$. Combine that into a single exponential.
Notice how that integral is over a length of \$T\$, which is one period. It can only be non-zero for a single value.