Finding the Fourier Series of $\sin(x)^2\cos(x)^3$ I'm currently struggling at calculation the Fourier series of the given function
$$\sin(x)^2 \cos(x)^3$$
Given Euler's identity, I thought that using the exponential approach would be the easiest way to do it.
What I found was: $$\frac{-1}{32}((\exp(2ix)-2\exp(2ix)+\exp(-2ix))(\exp(3ix)+3\exp(ix)+3\exp(-ix)+\exp(-3ix)))$$
Transforming it back, the result is:
$$ -\frac{1}{18}(\cos(5x)+\cos(3x)+2\cos(x))$$
(I've checked my calculations multiple times, I'm pretty sure it's correct.)
Considering the point $x = 0$ however, one can see that the series I found doesn't match the original function.
Could someone help me find my mistake?

1) Trigonometric identities:
$$
\sin^2 x\cos^3x=(\sin x\cos x)^2\cos x=\left(\frac{\sin 2x}{2}\right)^2\cos x=\frac{1}{4}\sin^22x\cos x
$$
$$
=\frac{1}{4}\left(\frac{1-\cos 4x}{2}\right)\cos x=\frac{\cos x}{8}-\frac{\cos 4x\cos x}{8}
$$
$$
=\frac{\cos x}{8}-\frac{\cos 5x+\cos 3x}{16}
$$
$$
=\frac{\cos x}{8}-\frac{\cos 3x}{16}-\frac{\cos 5x}{16}
$$
2) Complex exponential:
$$
\sin^2x\cos^3x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2\left(\frac{e^{ix}+e^{-ix}}{2}\right)^3
$$
$$
=-\frac{1}{32}(e^{2ix}-2+e^{-2ix})(e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix})
$$
$$
=-\frac{1}{32}(e^{5ix}+e^{3ix}-2e^{ix}-2e^{-ix}+e^{-3ix}+e^{-5ix})
$$
$$
=-\frac{1}{32}(2\cos 5x+2\cos 3x-4\cos x)
$$
$$
=\frac{1}{16}(2\cos x-\cos 3x-\cos 5x)
$$
Note: you made a mistake when you expanded $(e^{ix}-e^{-ix})^2$. I have no idea how you ended up with this $18$. You probably meant $16$.
