Why my Fourier coefficients are all zeros?

Question

given a function $f(x) = \frac{\pi -x}{2}$ on $(0,2\pi)$

then

$$a_n = \frac{1}{2\pi} \int_0^{2\pi} (\pi-x) cos(nx) dx = \frac{1}{2n\pi} \Big[\pi sin(nx) -\frac{1}{n}\Big(nxsin(nx) + cos(nx) \Big)\Big]_0^{2\pi}=0$$ $$b_n = \frac{1}{2\pi} \int_0^{2\pi} (\pi-x) sin(nx) dx = \frac{1}{2 n \pi} \Big[ -\pi cos(nx)-\frac{1}{n} \Big(-nxcos(nx)+sin(nx)\Big) \Big]_0^{2\pi}= 0$$

what am i doing wrong?

Answer 1

Here is a more detailed integration with limits plugged in. Note that $\sin(0) = \sin(2\pi n) = 0$ and $\cos(2\pi n) = 1$. $$ \begin{split} \int_0^{2\pi} (\pi-x) \sin(n x) dx &= \left.-\frac{\sin(n x)+n (\pi-x) \cos(n x)}{n^2} \right|_0^{2\pi} \\ &= \left.\frac{\sin(n x)+n (\pi-x) \cos(n x)}{n^2} \right|_{2\pi}^0 \\ &= \frac{n \pi}{n^2} - \frac{-n \pi}{n^2} = \frac{2\pi n}{n^2} = \frac{2\pi}{n}. \end{split} $$

Answer 2

$b_n = \frac{1}{2\pi n} [xcos(nx)]_0^{2\pi} = \frac{1}{n}$