$ Let\\[2ex] f\left(x\right)=\ x^2=a_0+\sum_{n=1}^{\infty{}}\left(a_n\cos{nx}+b_n\sin{nx}\right)\ \ \ \ \ \ \ \ \ ......\ (1)\\[5ex] Now,\\[2ex] \begin{align*} \displaystyle a_0\ &=\ \frac{1}{2\pi{}}\int_0^{2\pi{}}f\left(x\right)\ dx\ \\[2ex] \displaystyle &=\frac{1}{2\pi{}}\int_0^{2\pi{}}{\ x}^2dx \\[2ex] \displaystyle &=\frac{1}{2\pi{}}{\left[\frac{{\ x}^3}{3}\right]}_0^{2\pi{}} \\[2ex] \displaystyle \\[2ex] \displaystyle &=\ \frac{4{\ \pi{}}^2}{3} \\[2ex] \displaystyle a_n&=\ \frac{1}{\pi{}}\int_0^{2\pi{}}f\left(x\right)\cos nx\ dx\ \\[2ex] \displaystyle &=\ \frac{1}{\pi{}}\int_0^{2\pi{}}{\ x}^2\cos nx\ dx\ \\[6ex] By\ & integration \ by \ parts,\\[4ex] \displaystyle &= \frac{1}{\pi{}}{\left[{\ x}^2\frac{\sin {nx}}{n}-\ 2x\left(-\frac{\cos {nx}}{{\ n}^2}\right)+\ 2\left(-\frac{\sin{nx}}{{\ n}^3}\right)\right]}_0^{2\pi{}} \\[2ex] \displaystyle &=\ \frac{1}{\pi{}}\left[\frac{4\pi{}}{{\ n}^2}\right] \\[2ex] \displaystyle &=\frac{4}{{\ n}^2} \\[2ex] \displaystyle b_n&=\ \frac{1}{\pi{}}\int_0^{2\pi{}}f\left(x\right)\sin nx\ dx\ \\[2ex] \displaystyle &=\ \frac{1}{\pi{}}\int_0^{2\pi{}}{\ x}^2\sin nx\ dx\ \\[6ex] By & \ integration \ by \ parts,\\[4ex] \displaystyle &= \frac{1}{\pi{}}{\left[{\ x}^2\left(-\frac{\cos{nx}}{n}\right)-\ 2x\left(-\frac{\sin{nx}}{{\ n}^2}\right)+\ 2\left(\frac{\cos{nx}}{{\ n}^3}\right)\right]}_0^{2\pi{}} \\[2ex] \displaystyle &=\frac{1}{\ \pi{}}\ \left[\left(-\frac{4{\pi{}}^2}{n}+\ \frac{2}{{\ n}^3}\right)-\ \frac{2}{{\ n}^3}\right] \\[2ex] \displaystyle &=\ \frac{4\pi{}}{n} \\[2ex] \end{align*}\\ $

$Substituting \ this \ values \ in \ (1)\\[2ex] \displaystyle x^2=\frac{4{\ \pi{}}^2}{3}+\sum_{n=1}^{\infty{}}\frac{4}{{\ n}^2}\cos{nx-}\sum_{n=1}^{\infty{}}\frac{4\pi{}}{n}\sin{nx} \\[2ex] \displaystyle x^2=\frac{4{\ \pi{}}^2}{3}+4\sum_{n=1}^{\infty{}}\frac{1}{{\ n}^2}\cos{nx-4\pi{}}\sum_{n=1}^{\infty{}}\frac{1}{n}\sin{nx} \\[2ex] $