Half range sine series is given by

$b_{n} = \frac{2}{l}\int\limits_{0}^{l}f(x)sin\bigg(\frac{n\pi x}{l}\bigg) dx$ a = 0 & $a_{n}$ = 0

Here $f(x) = x^2$ and l = 3

$b_{n} = \frac{2}{3}\int\limits_{0}^{3}x^2sin\bigg(\frac{n\pi x}{3}\bigg) dx$

Using Leibnitz rule,

$b_{n} = \frac{2}{3}\bigg[x^2 \frac{\bigg(\frac{-cosn\pi x}{3}\bigg)}{\frac{n\pi}{3}} - \frac{(2x)\frac{-sinn\pi x}{3}}{\bigg(\frac{n\pi}{3}\bigg)^2} + \frac{2\bigg(\frac{cosn\pi x}{3}\bigg)}{\bigg(\frac{n\pi}{3}\bigg)^3}\bigg]_{0}^{3}$

$b_{n} = \frac{2}{3}\bigg[3^2 \frac{3}{n\pi}(-cosn\pi) - 6\frac{9}{n^2 \pi^2}(-sinn\pi) + 2\frac{3^3}{n^3 \pi^3}cosn\pi - 2\frac{3^3}{n^3 \pi^3}\bigg]$

$b_{n} = \frac{2}{3}\bigg[\frac{3^3}{n\pi}(-cosn \pi) - 0 + 2 \frac{3^3}{n^3 \pi^3}cosn\pi - 2\frac{3^3}{n^3 \pi^3}\bigg]$

$b_{n} = \frac{2}{3}\bigg[\frac{3^3}{n\pi} - \frac{2*2*3^3}{n^3 \pi^3}\bigg]$ If n is odd

$b_{n} = \frac{2}{3}\bigg[-\frac{3^3}{n\pi}\bigg]$ if n is even

i.e.

$b_{n} = \frac{2}{\pi}\bigg[\frac{9}{n} - \frac{6}{n^3 \pi^3}\bigg]$ if n is odd

$b_{n} = \frac{2}{\pi}\bigg[-\frac{9}{n}\bigg]$ if n is even

$\therefore x^2 = \frac{2}{\pi}\bigg(9 - \frac{36}{\pi^2}\bigg)sinx - \frac{9}{2}sin2x + \bigg(9 - \frac{36}{3^3 \pi^2}\bigg)sin3x.......$