$\displaystyle = \frac{1}{\pi{}}\ {\left[\frac{\sin h3x}{3}\right]}_0^{\pi{}} \\[2ex] \displaystyle = \frac{\sin h3\pi{}}{3\pi{}} \\[2ex] \displaystyle a_n = \frac{2}{\pi{}}\ \int_0^{\pi{}}f\left(x\right)\cos {nx}\ dx \\[2ex] \displaystyle = \frac{2}{\pi{}}\ \int_0^{\pi{}}\cos h3x\cos {nx}\ dx \\[2ex] \displaystyle = \frac{2}{\pi{}} \int_0^{\pi{}}\frac{e^{3x}+e^{-3x}}{2}\cos {nx}\ dx \\[2ex] \displaystyle = \frac{1}{\pi{}}\ {\left[\frac{e^{3x}}{3^2+n^2}\ \left(3\cos nx+n\sin nx\right)+\frac{e^{-3x}}{3^2+n^2} \left(-3\cos nx+n\sin nx\right)\right]}_0^{\pi{}} \\[2ex] \displaystyle =\frac{1}{\pi{}}\ \left[\frac{e^{3\pi{}}}{3^2+n^2}\ \left(3\cos n\pi{}\right)-\frac{e^{-3\pi{}}}{3^2+n^2}\ \left(3\cos n\pi{}\right)-\frac{1}{3^2+n^2}\left(3\right)+\frac{1}{3^2+n^2}\left(3\right)\right] \\[2ex] \displaystyle = \frac{1}{\pi{}}\ {\left[\frac{\sin h3x}{3}\right]}_0^{\pi{}} \\[2ex] \displaystyle = \frac{\sin h3\pi{}}{3\pi{}} \\[2ex] \displaystyle a_n = \frac{2}{\pi{}}\ \int_0^{\pi{}}f\left(x\right)\cos {nx}\ dx \\[2ex] \displaystyle = \frac{2}{\pi{}}\ \int_0^{\pi{}}\cos h3x\cos {nx}\ dx \\[2ex] \displaystyle = \frac{2}{\pi{}}\int_0^{\pi{}}\frac{e^{3x}+e^{-3x}}{2}\cos {nx}\ dx \\[2ex] \displaystyle = \frac{1}{\pi{}}\ {\left[\frac{e^{3x}}{3^2+n^2}\ \left(3\cos nx+n\sin nx\right)+\frac{e^{-3x}}{3^2+n^2} \left(-3\cos nx+n\sin nx\right)\right]}_0^{\pi{}} \\[2ex] \displaystyle =\frac{1}{\pi{}}\ \left[\frac{e^{3\pi{}}}{3^2+n^2}\ \left(3\cos n\pi{}\right)-\frac{e^{-3\pi{}}}{3^2+n^2}\ \left(3\cos n\pi{}\right)-\frac{1}{3^2+n^2}\left(3\right)+\frac{1}{3^2+n^2}\left(3\right)\right] \\[2ex] \displaystyle = \frac{3{(-1)}^n}{\pi{}\left(9+n^2\right)} [e^{3\pi{}}-e^{-3\pi{}}] \\[2ex] \displaystyle = \frac{6{(-1)}^n}{\pi{}\left(9+n^2\right)} \sin h3\pi{} \\[2ex] {\ f}_2\left(x\ \right)\ is\ an\ odd\ function \\[2ex] \therefore{}\ a_0=0\ ,\ a_n=0 \\[2ex] \displaystyle {\ f}_2\left(x\ \right)=\sum_{n=1}^{\infty{}}\left(b_n\sin {nx}\right) \\[2ex] \displaystyle b_n = \frac{2}{\pi{}}\ \int_0^{\pi{}}{\ f}_2\left(x\ \right)\sin nx\ dx\ \\[2ex] \displaystyle = \frac{2}{\pi{}}\ \int_0^{\pi{}}\sin h3x\ \sin nx\ dx\ \\[2ex] \displaystyle = \frac{2}{\pi{}}\ \int_0^{\pi{}}\frac{e^{3x}\ -\ e^{-3x}}{2}\ \sin nx\ dx\ \\[2ex] $