Answer:

$\begin{align*} f\left(x\right)&=a_0+\sum_{n=1}^{\infty{}}\left(a_n\cos{nx}+b_n\sin{nx}\right)\ \ \ \ \ ......(1) \\[2ex] \displaystyle a_0\ &=\ \frac{1}{2\pi{}}\int_{-\pi{}}^{\pi{}}f\left(x\right)\ dx\ \\[2ex] \displaystyle &=\ \frac{1}{2\pi{}}\left[\int_{-\pi{}}^0x\ dx+\ \ \int_0^{\frac{\pi{}}{2}}0\ dx+\ \int_{\frac{\pi{}}{2}}^{\pi{}}\left(x-\frac{\pi{}}{2}\right)\ dx\ \ \right] \\[2ex] \displaystyle &=\ \frac{1}{2\pi{}}\left[{\left(\frac{x^2}{2}\right)}_{-\pi{}}^0+\ 0+\ {\left(\frac{x^2}{2}-\frac{\pi{}}{2}x\right)}_{\frac{\pi{}}{2}}^{\pi{}}\right] \\[2ex] \displaystyle &=\ \frac{1}{2\pi{}}\left[\left(0-\frac{{\pi{}}^2}{2}\right)+\ \ \left(\frac{{\pi{}}^2}{2}-\frac{\pi{}}{2}\pi{}\right)-\ \left(\frac{1}{2}\frac{{\pi{}}^2}{2^2}-\frac{\pi{}}{2}\frac{\pi{}}{2}\right)\right] \\[2ex] \displaystyle &=\ \frac{1}{2\pi{}}\left[\left(-\frac{{\pi{}}^2}{2}\right)+0-\ \ \left(\frac{{\pi{}}^2}{8}\right)+\ \left(\frac{{\pi{}}^2}{4}\right)\right] \\[2ex] \displaystyle &=\frac{1}{2\pi{}}\left[\left(-\frac{{3\pi{}}^2}{8}\right)\right] \\[2ex] \displaystyle &=\ -\frac{3\pi{}}{16} \\[2ex] \displaystyle a_n&=\ \ \frac{1}{\pi{}}\int_{-\pi{}}^{\pi{}}f\left(x\right)\cos nx\ dx\ \\[2ex] \displaystyle &=\ \frac{1}{\pi{}}\left[\int_{-\pi{}}^0x \cos\ nx\ dx+\ \ \int_0^{\frac{\pi{}}{2}}0\ (\cos nx)dx+\ \int_{\frac{\pi{}}{2}}^{\pi{}}\left(x-\frac{\pi{}}{2}\right)\cos {nx}\ dx\ \ \right] \\[2ex] \displaystyle &=\ \frac{1}{\pi{}}[{\left(x\frac{\sin{nx}}{n}-\left(1\right)\left(\frac{-\cos{nx}}{n^2}\right)\right)}_{-\pi{}}^0+\ 0 \\[2ex] & \ \ \ \ \ \ \ +\ {\left(\left(x-\frac{\pi{}}{2}\right)\frac{\sin{nx}}{n}-\left(1\right)\left(\frac{-\cos{nx}}{n^2}\right)\right)}_{\frac{\pi{}}{2}}^{\pi{}}] \\[2ex] \displaystyle &=\ \frac{1}{\pi{}}\left[0+\frac{1}{n^2}-0-\frac{\cos{\left(-n\pi{}\right)}}{n^2}+0+\frac{\cos{\left(n\pi{}\right)}}{n^2}-0-\frac{\cos{\left(n\frac{\pi{}}{2}\right)}}{n^2}\right] \\[2ex] \displaystyle &=\ \frac{1}{\pi{}}\left[\frac{1}{n^2}-\frac{\cos{\left(n\pi{}\right)}}{n^2}+\frac{\cos{\left(n\pi{}\right)}}{n^2}-\frac{\cos{\left(n\frac{\pi{}}{2}\right)}}{n^2}\right] \\[2ex] \displaystyle &=\ \frac{1}{n^2\pi{}}\left[1-\ \cos{\left(n\frac{\pi{}}{2}\right)}\right] \\[2ex] \displaystyle b_n&=\ \frac{1}{\pi{}}\int_{-\pi{}}^{\pi{}}f\left(x\right)\sin nx\ dx\ \\[2ex] \displaystyle &=\ \frac{1}{\pi{}}\left[\int_{-\pi{}}^0xs\in nx\ dx+\ \ \int_0^{\frac{\pi{}}{2}}0\ (\sin nx)dx+\ \int_{\frac{\pi{}}{2}}^{\pi{}}\left(x-\frac{\pi{}}{2}\right)\sin{nx}\ dx\ \ \right] \\[2ex] \displaystyle &=\ \frac{1}{\pi{}} \Bigg[{\left(x\frac{-\cos{nx}}{n}-\left(1\right)\left(\frac{-\sin{nx}}{n^2}\right)\right)}_{-\pi{}}^0+\ 0\\[2ex]& \ \ \ \ \ \ \ \ +\ {\left(\left(x-\frac{\pi{}}{2}\right)\frac{-\cos{nx}}{n}-\left(1\right)\left(\frac{-\sin{nx}}{n^2}\right)\right)}_{\frac{\pi{}}{2}}^{\pi{}} \Bigg] \\[2ex] \displaystyle &=\frac{1}{\pi{}}\left[-0+0+\left(-\pi{}\right)\frac{\cos{n\pi{}}}{n}-0-\left(\pi{}-\frac{\pi{}}{2}\right)\frac{\cos{n\pi{}}}{n}+0+0-\frac{\sin{\left(n\frac{\pi{}}{2}\right)}}{n^2}\right] \\[2ex] \displaystyle &=\ \frac{1}{\pi{}}\left[\left(-\pi{}\right)\frac{\cos{n\pi{}}}{n}-\frac{\pi{}}{2}\frac{\cos{n\pi{}}}{n}-\frac{\sin{\left(n\frac{\pi{}}{2}\right)}}{n^2}\right] \\[2ex] \displaystyle &=\ \frac{-1}{\pi{}}\left[\frac{3\pi{}}{2}\frac{\cos{n\pi{}}}{n}+\frac{\sin{\left(n\frac{\pi{}}{2}\right)}}{n^2}\right] \\[2ex] \end{align*}\\[3ex] Substituting \ in \ equation \\[3ex] \displaystyle f\left(x\right)=-\frac{3\pi{}}{16}+\frac{1}{\pi{}}\sum_{n=1}^{\infty{}}\frac{1}{n^2}\left[1-\ \cos{\left(n\frac{\pi{}}{2}\right)}\right]\left(\cos{nx}\right)\\[2ex] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \displaystyle -\frac{1}{\pi{}}\sum_{n=1}^{\infty{}}\left[\frac{3\pi{}}{2}\frac{\cos{n\pi{}}}{n}+\frac{\sin{\left(n\frac{\pi{}}{2}\right)}}{n^2}\right]\left(\sin{nx}\right) \\[2ex] $