$\displaystyle For\ half\ range\ cosine\ series\ b_n=0 \\[2ex] \displaystyle a_0=\frac{2}{l}\int_0^lf\left(x\right)dx \\[2ex] \displaystyle =\frac{2}{l}\int_0^2x\ dx \\[2ex] \displaystyle =\frac{2}{l}{\left[\frac{x^2}{2}\right]}_0^2=\frac{2}{l}\times{}\frac{4}{2}=\frac{4}{l} \\[2ex] $

$\displaystyle a_n=\frac{2}{l}\int_0^lf\left(x\right)\cos{\frac{n\pi{}x}{l}dx} \\[2ex] \displaystyle =\frac{2}{l}\int_0^l\ x\cos{\frac{n\pi{}x}{l}dx} \\[2ex] \displaystyle =\frac{2}{l}{\left[x\sin{\frac{n\pi{}x}{l}}\bullet{}\frac{l}{n\pi{}}-1\left[-cos\frac{n\pi{}x}{l}\bullet{}{\left(\frac{l}{n\pi{}}\right)}^2\right]\right]}_0^{l\ } \\[2ex] \displaystyle =\frac{2}{l}\left[l\times{}0+\cos n\pi{}\times{}{\left(\frac{l}{n\pi{}}\right)}^2-0+\ {\left(\frac{l}{n\pi{}}\right)}^2\right] \\[2ex] \displaystyle \left\{\sin{n\pi{}=0,\ }\cos{n\pi{}}={\left(-1\right)}^n\right\} \\[2ex] \displaystyle =\frac{2l}{n^2{\pi{}}^2}[{\left(-1\right)}^n+1] \\[2ex] $

$\displaystyle \therefore{}Half\ range\ fourier\ cosine\ series\ is \\[2ex] \displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty{}}an\ \cos{\frac{n\pi{}x}{l}} \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}x=\frac{2}{l}+\sum_{n=1}^{\infty{}}\frac{2l[{\left(-1\right)}^n+1]}{n^2{\pi{}}^2}\cos{\frac{n\pi{}x}{l}} \\[2ex] \displaystyle x=\frac{2}{l}-\frac{2l}{{\pi{}}^2}\left[\frac{1}{2^2}\cos{\frac{2\pi{}x}{l}+\frac{1}{4^2}\cos{\frac{4\pi{}x}{l}}}+…\right] \\[2ex] $