The fourier series is given by

$f(x) = \frac{a_{0}}{2} + \sum\limits_{n = 1}^{\infty}(a_{n}cosnx + b_{n}sinnx)$

Here the period is 2$\pi$

Let us first find the coefficients $a_{0}, a_{n}$ & $b_{n}$

$\therefore a_{0} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)dx$

$\therefore a_{0} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)dx$

$\therefore a_{0} = \frac{1}{\pi}\int\limits_{-\pi}^{0}0dx + \int\limits_{0}^{\pi}x^2dx$

$\therefore a_{0} = \frac{1}{\pi}\bigg[\frac{x^3}{3}\bigg]_{0}^{\pi} = \frac{\pi^2}{2}$

$\therefore a_{0} = \frac{\pi^2}{2}$

Now,

$a_{n} = \frac{1}{\pi}\int\limits_{-\pi}^{\pi}f(x)cosnxdx$

$\therefore a_{n} = \frac{1}{\pi}\bigg[\int\limits_{-\pi}^{0}0*cosnxdx + \int\limits_{0}^{\pi}x^2 cosnxdx\bigg]$

$\therefore a_{n} = \frac{1}{\pi}\bigg[x^2 \frac{sinnx}{n} - 2x\bigg(\frac{-cosnx}{n^2}\bigg) + 2\bigg(\frac{-sinnx}{n^3}\bigg)\bigg]_{0}^{\pi}$

[using Leibnitz rule]

$a_{n} = \frac{1}{\pi}\bigg[\pi^2 \frac{sinn\pi}{n} - 2\pi\bigg]\bigg(\frac{-cosn\pi}{n^2}\bigg) + \bigg(\frac{-sinn\pi}{n^3} - 0 + 0 - 0\bigg)$

$a_{n} = \frac{1}{\pi}\bigg[\frac{2\pi cosn\pi}{n^2}\bigg]$

$a_{n} = \frac{2}{n^2}(-1)^n..........n = 1,2,3$

Now

$b_{n} = \frac{1}{\pi} \int\limits_{-\pi}^{\pi}f(x)sinnx dx$

$b_{n} = \frac{1}{\pi}\int\limits_{-pi}^{0}0*sinnxdx + \int\limits_{0}^{\pi}x^2sinnxdx$

$b_{n} = \frac{1}{\pi}\bigg[x^2\bigg(\frac{-cosnx}{n}\bigg) - 2x\bigg(\frac{-sinx}{n^2}\bigg) + 2\bigg(\frac{cosnx}{n^3}\bigg)\bigg]_{0}^{\pi}$

$b_{n} = \frac{1}{\pi}\bigg[\pi^2\bigg(\frac{-cosn\pi}{n}\bigg) - 2x\bigg(\frac{-sin\pi}{n^2}\bigg) + 2\bigg(\frac{cosn\pi}{n^3}\bigg) - 0 + 0 - \frac{2}{n^3}\bigg]$

$b_{n} = \frac{1}{\pi}\bigg[\pi^2 \bigg(\frac{(-1)^{n + 1}}{n}\bigg) + \frac{2(-1)^n}{n^3} - \frac{2}{n^3}\bigg]$

$b_{n} = \frac{1}{\pi}\bigg[\pi^2 \bigg(\frac{(-1)^{n + 1}}{n}\bigg) + \frac{2(-1)^n}{n^3} - \frac{2}{n^3}\bigg] .........n = 1,2,3$

$f(x) = \frac{\pi^2}{6} + \bigg[\sum\limits_{n = 1}^{\infty} \frac{2}{n^2}(-`1)^n cosnx + \frac{1}{\pi} \sum\limits_{n = 1}^{\infty} \frac{2}{n^2}(-1)^n cosnx + \frac{1}{\pi} \sum\limits_{n=1}^{\infty} \frac{\pi^2 (-1)^{n + 1}}{n} + \frac{2(-1)^{n + 1}}{n^3}\bigg]sinx$