$\displaystyle Let\ c=0\ and\ c+l=2\pi{}. \\[2ex] \displaystyle \therefore{}l=\pi{}. \\[2ex] $

$\displaystyle Now,\\[2ex] \displaystyle a_0=\frac{1}{l}\int_c^{c+2l}f\left(x\right)dx \\[2ex] \displaystyle ={\frac{1}{\pi{}}\int_0^{2\pi{}}xdx=\frac{1}{\pi{}}\left[\frac{{\ x}^2}{2}\right]}_0^{2\pi{}} \\[2ex] \displaystyle =\frac{1}{2\pi{}}\ [(2{\pi{})}^2-0]=2\pi{} \\[2ex] $

$\displaystyle \therefore{}a_n=\ \frac{1}{l}\int_c^{c+2l}f\left(x\right)\cos\frac{n\pi{}x}{l}dx \\[2ex] \displaystyle =\frac{1}{\pi{}}\int_0^{2\pi{}}x\cos\frac{n\pi{}x}{l}dx \\[2ex] \displaystyle =\frac{1}{\pi{}}\int_0^{2\pi{}}x\cos\frac{n\pi{}x}{l}dx \\[2ex] \displaystyle =\frac{1}{\pi{}}{\left[x.\frac{\sin\ nx}{n}-1.\frac{-\cos\ nx}{{\ n}^2}\right]}_0^{2\pi{}} \\[2ex] \displaystyle =\frac{1}{\pi{}}\left[\left(2\pi{}.\frac{\sin\ 2\pi{}n}{n}+\frac{\cos\ 2n\pi{}}{{\ n}^2}\right)-\left(0+\frac{\cos\ 0}{{\ n}^2}\right)\right]\ \\[2ex] \displaystyle =\frac{1}{\pi{}}\left[0+\frac{1}{{\ n}^2}-0-\frac{1}{{\ n}^2}\right]=0 \\[2ex] $

$\displaystyle \therefore{}b_n=\ \frac{1}{l}\int_c^{c+2l}f\left(x\right)\sin\frac{n\pi{}x}{l}dx \\[2ex] \displaystyle =\frac{1}{\pi{}}\int_0^{2\pi{}}x\sin\frac{n\pi{}x}{l}dx \\[2ex] \displaystyle =\frac{1}{\pi{}}{\left[x.\frac{-\cos\ nx}{n}-1.\frac{-\sin\ nx}{{\ n}^2}\right]}_0^{2\pi{}} \\[2ex] \displaystyle =\frac{1}{\pi{}}\left[\left(2\pi{}.\frac{-\cos\ 2\pi{}n}{n}+\frac{\sin\ 2n\pi{}}{{\ n}^2}\right)-\left(0+\frac{\sin\ 0}{{\ n}^2}\right)\right]\ \\[2ex] \displaystyle =\frac{1}{\pi{}}\left[-\frac{2\pi{}}{{\ n}^1}+0-0-0\right]=-\frac{2}{n} \\[2ex] $

$\displaystyle In\ fourier\ series, \\[2ex] \displaystyle f(x)=\ \frac{a_0}{2}+\sum_{n=1}^{\infty{}}a_n\ \cos\frac{n\pi{}x}{l}+\sum_{n=1}^{\infty{}}b_n\sin\frac{n\pi{}x}{l} \\[2ex] \displaystyle \therefore{}x=\pi{}+0+\sum_{n=1}^{\infty{}}-\frac{2}{n}\sin\frac{n\pi{}x}{\pi{}} \\[2ex] \displaystyle \therefore{}x=\pi{}-2\left[\frac{\sin1x}{1}+\frac{\sin2x}{2}+…\right] \\[2ex]$