If we're given the Fourier series of $e^x$ on the interval $(0,2\pi)$, I'm wondering if there's a nicer way to extract the sine series of $e^x$ on the same interval other than getting the coefficients by the integral $\int e^x sin(nx) dx$. For the record
$e^x = ((e^{2\pi}-1)/2\pi)\sum_{-\infty}^{\infty} \frac{cos(nx)+i sin(nx)}{1-in}$ on $(0,2\pi)$.
I am not quite sure why you might want another approach, seeing that the relevant integral is quite straightforward. But just for the record, here is an approach using a differential equation and assuming knowledge of the complex Fourier series of $x$ on the interval $[0,2\pi]$, which is $$ \pi + \sum_{\stackrel{n=-\infty}{n\not=0}}^\infty \frac{ie^{inx}}{n}\ . $$
Let's look at the function $$ f(x) = e^x - \frac{e^{2\pi}-1}{2\pi} x\ . $$ This is smooth on $(0,2\pi)$ and satisfies $f(0) = f(2\pi) (= 1)$ so we can differentiate its Fourier series term-by-term. And it satisfies the differential equation $$f' -f = \frac{e^{2\pi}-1}{2\pi} (x-1)\ . $$ Suppose now the Fourier series of $f$ is $$ \sum_{n=-\infty}^\infty c_n e^{inx}\ . $$ Substituting into the differential equation we have $$ \sum_{n=-\infty}^\infty (in-1) c_n e^{inx} = \frac{e^{2\pi}-1}{2\pi} \left( \pi - 1 + \sum_{\stackrel{n=-\infty}{n\not=0}}^\infty \frac{ie^{inx}}{n} \right) $$ so $$ c_0 = \frac{(e^{2\pi}-1)(1-\pi)}{2\pi} \ , \quad c_n = \frac{e^{2\pi}-1}{2\pi}\frac{i}{n(in-1)} \quad n\not=0 $$ Finally we use $e^x = f + \frac{e^{2\pi}-1}{2\pi} x $ to get the Fourier series of $e^x$. It is $$ \left(\frac{(e^{2\pi}-1)(1-\pi)}{2\pi} + \sum_{\stackrel{n=-\infty}{n\not=0}}^\infty \frac{e^{2\pi}-1}{2\pi}\frac{i}{n(in-1)} e^{inx}\right) + \frac{e^{2\pi}-1}{2\pi}\left( \pi + \sum_{\stackrel{n=-\infty}{n\not=0}}^\infty \frac{ie^{inx}}{n} \right) $$ which simplifies to $$ \frac{e^{2\pi}-1}{2\pi} \sum_{n=-\infty}^\infty \frac{e^{inx}}{1-in} $$
