$\displaystyle f\left(x\right)=\vert{}\sin{x\vert{}\ ;\\[2ex]f\left(-x\right)=\vert{}\sin⁡(-x)\vert{}=\vert{}-\sin{x\vert{}=\sin{x=f(x)}}}. \\[2ex] \displaystyle \therefore{}\vert{}\sin{x\vert{}\ is\ even\ function.} \\[2ex] \displaystyle \therefore{}b_n=0 \\[2ex] $

$\displaystyle In\ \left(0,2\pi{}\right)\ i.e\ \left(1st\ and\ 2nd\ quadrant\right),\ Sin\ is\ positive.\ \\[2ex] \displaystyle \therefore{}f\left(x\right)=\vert{}\sin{x\vert{}=\sin{x\ }} \\[2ex] \displaystyle \therefore{}a_0=\frac{2}{\pi{}}\int_0^{\pi{}}\vert{}\sin{x\vert{}\ dx} \\[2ex] \displaystyle ={-\frac{2}{\pi{}}[-\cos{x]}}_0^{\pi{}} \\[2ex] \displaystyle =-\frac{2}{\pi{}}\left[-1-1\right]=\frac{4}{\pi{}} \\[2ex] $

$\displaystyle \therefore{}a_n=\frac{2}{\pi{}}\int_0^{\pi{}}\vert{}\sin{x\vert{}.\cos⁡(\frac{n\pi{}x}{\pi{}})\ dx} \\[2ex] \displaystyle =\frac{1}{\pi{}}\int_0^{\pi{}}2\cos nx.\sin xdx \\[2ex] \displaystyle =\frac{1}{\pi{}}\int_0^{\pi{}}\sin{\left(nx+x\right)}-\sin{\left(nx-x\right)}dx…\left(1\right) \\[2ex] \displaystyle ={\frac{1}{\pi{}}\left[-\frac{\cos{\left(n+1\right)}x}{n+1}+\frac{\cos{\left(n-1\right)}x}{n-1}\right]}_0^{\pi{}} \\[2ex] \displaystyle =\frac{1}{\pi{}}\left[\frac{{\left(-1\right)}^n}{n+1}-\frac{{\left(-1\right)}^n}{n-1}\right]-\left[-\frac{1}{n+1}-\frac{1}{n-1}\right] \\[2ex] \displaystyle =\frac{1}{\pi{}}\left[\frac{1}{n+1}-\frac{1}{n-1}\right].[{\left(-1\right)}^n+1] \\[2ex] \displaystyle =\frac{{\left(-1\right)}^n+1}{\pi{}}\left[\frac{n-1-n-1}{\left(n+1\right)\left(n-1\right)}\right]....(2)\\[2ex] $

$Put\ n=1\ in\ (2) \\[2ex] \displaystyle \therefore{}a_1=\frac{1}{\pi{}}\int_0^{\pi{}}\sin{2x-0} \\[2ex] \displaystyle =\frac{1}{\pi{}}{\left[\frac{-\cos{2x}}{2}\right]}_0^{\pi{}} \\[2ex] \displaystyle =\frac{1}{\pi{}}\times{}-\frac{1}{2}\left[1-1\right]=0 \\[2ex] $

$\displaystyle By\ fourier\ series, \\[2ex] \displaystyle f\left(x\right)=\frac{2}{\pi{}}+0+\sum_2^{\infty{}}\frac{{\left(-1\right)}^n+1}{\pi{}}\left[\frac{n-1-n-1}{\left(n+1\right)\left(n-1\right)}\right]\cos nx \\[2ex] \displaystyle \vert{}\sin{x\vert{}=\frac{2}{\pi{}}+\frac{1}{\pi{}}\left[2\left(\frac{1}{3}-\frac{1}{1}\right)\cos{2x+0+2\left(\frac{1}{5}-\frac{1}{3}\right)\cos{4x}}\right]} \\[2ex] \displaystyle \vert{}\sin{x\vert{}=\frac{2}{\pi{}}+\frac{2}{\pi{}}\left[\left(\frac{1}{3}-\frac{1}{1}\right)\cos{2x+\left(\frac{1}{5}-\frac{1}{3}\right)\cos{4x}+…}\right]} \\[2ex]$