Fourier series representation of $\sin^4 x$ I tried solving for fourier coefficients of Fourier series for the multiples of fundamental frequency $\omega_0=2$. So $F_n=\int_0^{\pi} \sin^4 x \, e^{-i2nx} dx$. And my calculator says answer should be $3e^{-i2nx}/8$.
But for some reasons I get $i(2+1/(4-2n)-1/(4+2n))/(16\pi)$. Can you show how one gets the correct answer without using trigonometry and just using integration of complex exponentials?

Simply rewrite $\sin^4 x$ in terms of multiple angle trigonometric functions. You will get a finite series.
In fact 
$$
1 - 2 \sin^2 x = \cos(2x) \quad \text{and} \quad \cos(2x) = 2\cos^2(x)-1\\
\Rightarrow \sin^2 x = \frac{1-\cos(2x)}{2} \\
\Rightarrow \sin^4 x = \frac{1+\cos^2(2x)-2\cos(2x)}{4} \\
\Rightarrow \sin^4 x = \frac{1+(\cos(4x)+1)/2-2\cos(2x)}{4} \\
= \frac{3}{8} - \frac{\cos(2x)}{2} + \frac{\cos(4x)}{8}  
$$
