So my TA in class introduced this amazing way of finding fourier series coefficients for a sin wave, by writing
$ sin( \omega t ) = (e^{i\omega t}-e^{-i\omega t}) / 2i $ ----(1)
Hence getting the form of the desired fourier coefficients.
Then I thought of substituting the same in the common formula given for fourier series
$x(t) = \sum_{k=-\infty}^{\infty}C_k e^{ik\omega _o t}, \omega_o=2\pi/(period)$ ----(2)
$C_n=\frac{1}{T} \int_T x(t)e^{-in\omega_o t}dt $ ----(3)
Putting RHS of (1) in (3)
$ \frac{\omega}{2\pi} \int _0^{2\pi} (e^{i\omega t}-e^{-i\omega t})e^{-in\omega t}dt $
Taking $f=i \omega t$ therefore $dt=\frac{df}{i\omega}$, Hence the equation becomes
$\frac{-1}{4\pi}\int_0^{i2\pi\omega} (e^f-e^{-f}) e^{-nf} df$
$\frac{-1}{4\pi}\int_0^{i2\pi\omega} (e^{f(1-n)}-e^{-f(1+n)})df$
Solving this gives and using that $e^{i2\pi}=1$ we get
$\frac{-1}{4\pi} [ \frac{1^{\omega (1-n)} -1}{1-n} + \frac{1^{\omega (1+n)} -1}{1+n} ] $
This conveniently gives 0 for all values except for n=-1 and n=1. How to evaluate this to match it with the coefficients of equation (1). I am also unable to calculate the limit, which I was getting as 0.