Find half range sine series for f(x)= πx-x2 in (0, π) Hence deduce that [\frac{{\pi{}}^3}{32}=\frac{1}{12}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\pi ]

$\displaystyle For\ half\ range\ sine\ series\ a_n=a_0=0\ \\[2ex] \displaystyle here,\ f\left(x\right)=x\ \left(\pi{}-x\right),\ l=\pi{} \\[2ex] \displaystyle b_n=\frac{2}{l}\int_0^lf\left(x\right)\sin{\frac{n\pi{}x}{l}dx} \\[2ex] \displaystyle =\frac{2}{\pi{}}\int_0^{\pi{}}x\left(\pi{}-x\right)\sin{nx}dx\ \ \\[2ex] \displaystyle =\frac{2}{\pi{}}{\left[x\left(\pi{}-x\right)\ \left(-\frac{\cos{nx}}{n}\right)-\left(\pi{}-2x\right)\left(-\frac{\sin{nx}}{n^2}\right)\ +\left(-2\right)\left(\frac{\cos{nx}}{n^3}\right)\right]}_0^{\pi{}} \\[2ex] \displaystyle =\frac{2}{\pi{}}\left[0-0-\frac{2{\left(-1\right)}^n}{n^3}+\frac{2}{n^3}\right]\ \\[2ex] \displaystyle \left\{\sin{n\pi{}}=0,\cos{n\pi{}}={\left(-1\right)}^n\right\}\ \\[2ex] $

$\displaystyle b_n=\frac{4}{\pi{}n^3}\ \left[1-{\left(-1\right)}^n\right] \\[2ex] \displaystyle \ \ \\[2ex] \displaystyle \therefore{}Half\ range\ fourier\ sine\ series\ is \\[2ex] \displaystyle \\[2ex] \displaystyle f(x)=\sum_{n=1}^{\infty{}}b_n\ \sin{\frac{n\pi{}x}{l}} \\[2ex] \displaystyle \ \ x\left(\pi{}-x\right)=\sum_{n=1}^{\infty{}}\ \frac{4}{\pi{}n^3}\ \left[1-{\left(-1\right)}^n\right]\sin{nx}\ \ \ \\[2ex] \displaystyle \ x\left(\pi{}-x\right)=\frac{4}{\pi{}}\sum_{n=1}^{\infty{}}\ \frac{\left[1-{\left(-1\right)}^n\right]}{n^3}\ \sin{nx}\ \ \ \\[2ex] $

$\displaystyle Put\ x=\frac{\pi{}}{2}\ ,\ \\[2ex] \displaystyle \frac{\pi{}}{2}\left(\frac{\pi{}}{2}\right)\ =\frac{4}{\pi{}}\ \left[\frac{2}{1^3}+0-\frac{2}{3^3}+0+\frac{2}{5^3}+0-\frac{2}{7^3}..\ \right]\ \\[2ex] \displaystyle \frac{{\pi{}}^3}{32}=\frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+..…\ \\[2ex] $