I have a function u(t) which is 0 for $ -\frac{\pi}{\omega} < t < 0 $ and $ Esin(\omega t)$ for $0 < t < \frac{\pi}{\omega}$. I calculated $a_0$ and $a_n$, but when I try to calculate $b_n = \frac{\omega E}{\pi} \int_0^\frac{\pi}{\omega} sin(\omega t)sin(n \omega t)dt$, I get the following expression which I don't know how to evaluate for $n=1$, and which is 0 for $n > 1$: $b_n = \frac{E}{2 \pi}(\frac{sin((1-n)\pi)}{1-n} - \frac{sin((1+n)\pi}{n+1})$
The answer is supposed to be $b_1 = \frac{E}{2}$. Could someone please explain how I can find this?
The general formula $$ b_n=\frac{E}{2\pi}\Bigl(\frac{\sin((1-n)\pi)}{1-n}-\frac{\sin((1+n)\pi)}{1+n}\Bigr) $$ is not valid for $n=1$, since then you divide by zero.
To calculate the coefficient $b_1$, you could reduce the integrand to $$ \sin^2(\omega t)=\frac{1-\cos(2\omega t)}{2}, $$ which you probably know how to integrate?