$(ii) L(\sin \ 2t \ \cos t \cos h2t) \\[2ex] \displaystyle \sin 2t \cos t = \frac{1}{2} [\sin 3t + \sin t] \\[2ex] \displaystyle \cos h2t = \frac{e^{2t}+e^{-2t}}{2} \\[2ex] \displaystyle \sin 2t \cos t \cos h2t = \frac{1}{4} [\sin 3t + \sin t]\left(\frac{e^{2t}+e^{-2t}}{2}\right) \\[2ex] \displaystyle = \frac{1}{4}\left[e^{2t}\left(\sin 3t\right)+e^{-2t}\left(\sin 3t\right)+e^{2t}\left(\sin t\right)+e^{-2t}\left(\sin t\right))\right] \\[2ex] \displaystyle \therefore{} L(\sin 2t \cos t \cos h2t) = \frac{1}{4} L\left[e^{2t}\left(\sin 3t\right)+e^{-2t}\left(\sin 3t\right)+e^{2t}\left(\sin t\right)+e^{-2t}\left(\sin t\right)\right] \\[2ex] \displaystyle L(\sin 3t) = \frac{3}{s^2{+3}^2} \\[2ex] \displaystyle L\left[e^{2t}\left(\sin 3t\right)+e^{-2t}\left(\sin 3t\right)\right]= \left[\frac{3}{{\left(s-2\right)}^2+9}+\frac{3}{{\left(s+2\right)}^2+9}\right] \\[2ex] \displaystyle = 3 \left[\frac{1}{s^2-4s+13}+\frac{1}{s^2+4s+13}\right] \\[2ex] \displaystyle = \frac{3.2(s^2+13)}{{s^4+10s}^2+{13}^2} \\[2ex] \displaystyle L(\sin t) = \frac{1}{s^2+1} \\[2ex] \displaystyle L\left[e^{2t}\left(\sin t\right)+e^{-2t}\left(\sin t\right)\right]= \left[\frac{1}{{\left(s-2\right)}^2+1}+\frac{1}{{\left(s+2\right)}^2+1}\right] \\[2ex] \displaystyle = \left[\frac{1}{s^2-4s+5}+\frac{1}{s^2+4s+5}\right] \\[2ex] \displaystyle = \frac{2(s^2+5)}{s^4-6s^2+5^2} \\[2ex] \displaystyle \therefore{} L(\sin 2t \cos t \cos h2t) = \frac{1}{4} \left[\frac{3.2(s^2+13)}{{s^4+10s}^2+{13}^2}+\frac{2(s^2+5)}{s^4-6s^2+5^2}\right] \\[2ex] \displaystyle = \frac{3(s^2+13)}{2\left({s^4+10s}^2+169\right)} + \frac{(s^2+5)}{2\left(s^4-6s^2+25\right)} \\[2ex] $