The Signal
Consider the signal $ 7\sin(2t)+(1+j)\cos(3t) $.
The Formulae
Recall the Fourier Series formulae for the continuous time signal case:
$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $
and
$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.
Finding the Series
It will be helpful -- necessary even -- to find the fundamental period of the signal. In our case, the period of the overall signal is $ 2\pi $, so $ \omega_0 $ will be $ \frac{2\pi}{2\pi}=1 $.
A good place to start is the calculation of $ a_0 $, which is the average of the signal. Plotting the signal makes it look like the average is 0, but we can integrate to check.
$ a_0=\frac{1}{T}\int_0^T[7\sin(2t)+(1+j)\cos(3t)]e^{-jk\omega_0t}dt $
$ =\frac{7}{2\pi}\int_0^{2\pi}\sin(2t)dt + \frac{1+j}{2\pi}\int_0^{2\pi}\cos(3t)dt $
$ =\frac{-7}{4\pi}\cos(2t)|_0^{2\pi}+\frac{1+j}{6\pi}\sin(3t)|_0^{2\pi} $
$ =\frac{-7}{4\pi}(\cos(4\pi)-\cos(0))+\frac{1+j}{6\pi}(\sin(6\pi)-\sin(0))=0 $
After this point, integrating becomes quite tedious, so I'll revert to using complex exponential identities to continue the solution. Our signal then becomes
$ x(t)=\frac{7}{2j}(e^{2j}-e^{-2j})+\frac{1+j}{2}(e^{3j}+e^{-3j}) $
$ =\frac{7}{2j}(e^2e^j-e^{-2}e^j)+\frac{1+j}{2}(e^3e^j+e^{-3}e^j) $
Now the Fourier coefficients should be fairly obvious.
$ a_{-3}=a_3=\frac{1+j}{2} $
$ a_{-2}\frac{-7}{2j} $
$ a_2\frac{7}{2j} $
All other $ a_k=0 $.
