Solution:

Let, $f(x)=\frac{1}{4}(\pi-x)^{2}=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n x+\sum_{n=1}^{\infty} b_{n}\sin nx,\\$ $By\ Euler's formula, we\ have$ $$ \begin{aligned} a_{0} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) dx \end{aligned} $$

$$=\frac{1}{\pi} \int_{0}^{2 \pi} \frac{1}{4}(\pi-x)^{2} d x$$

$$=\frac{1}{4 \pi}\left[\frac{(\pi-x)^{3}}{-3}\right]_{0}^{2 \pi}$$

$$=-\frac{1}{12 \pi}\left[-\pi^{3}-\pi^{3}\right]=\frac{\pi^{2}}{6}$$

$$ \begin{aligned} a_{n} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \cos n x d x=\frac{1}{\pi} \int_{0}^{2 \pi} \frac{1}{4}(\pi-x)^{2} \cos n x d x \\\\ &=\frac{1}{4 \pi}\left[(\pi-x)^{2} \frac{\sin n x}{n}-\{-2(\pi-x))\left(-\frac{\cos n x}{n^{2}}\right)+2\left(-\frac{\sin n x}{n^{3}}\right)\right]_{0}^{2 \pi} \\\\ &=\frac{1}{4 \pi}\left[\left(0+\frac{2 \pi \cos 2 n \pi}{n^{2}}+0\right)-\left(0-\frac{2 \pi \cos 0}{n^{2}}+0\right)\right]=\frac{1}{4 \pi}\left[\frac{2 \pi}{n^{2}}+\frac{2 \pi}{n^{2}}\right]=\frac{1}{n^{2}} \\\\ b_{n} &=\frac{1}{\pi} \int_{0}^{2 \pi} f(x) \sin n x d x=\frac{1}{\pi} \int_{0}^{2 \pi} \frac{1}{4}(\pi-x)^{2} \sin n x d x \\\\ &=\frac{1}{4 \pi}\left[(\pi-x)^{2}\left(-\frac{\cos n x}{n}\right)-[-2(\pi-x)\}\left(-\frac{\sin n x}{n^{2}}\right)+2\left(\frac{\cos n x}{n^{3}}\right)\right]_{0}^{2 \pi} \\\\ &=\frac{1}{4 \pi}\left[\left(-\frac{\pi^{2} \cos 2 n \pi}{n}-0+\frac{2 \cos 2 n \pi}{n^{3}}\right)-\left(-\frac{\pi^{2}}{n}-0+\frac{2 \cos 0}{n^{3}}\right)\right]\\\\ &=\frac{1}{4 \pi}\left[\left(-\frac{\pi^{2}}{n}+\frac{2}{n^{3}}\right)-\left(-\frac{\pi^{2}}{n}+\frac{2}{n^{3}}\right)\right]=0 \\ \end{aligned}\\ $$

$ \begin{aligned} \therefore \quad f(x) &=\frac{\pi^{2}}{12}+\sum_{n=1}^{\infty} \frac{\cos n x}{n^{2}}=\frac{\pi^{2}}{12}+\frac{\cos x}{1^{2}}+\frac{\cos 2 x}{2^{2}}+\frac{\cos 3 x}{3^{2}}+\ldots . . \end{aligned} $