i) Consider, 

$L[\sin 3u] = \dfrac {3}{s^2 +3^2}$

$ \therefore L [u\cdot \sin 3u] = (-1)^1 \dfrac {d}{ds} \dfrac {3}{s^2 +9} $

$= (-1) \times 3 \times \dfrac {-1}{(s^2 +9)^2} \cdot 2 s$

$= \dfrac {6s}{(s^2 +9)^2}$

$\displaystyle \therefore L \left [ \int^t_0 u \sin 3u \ du \right ] = \dfrac{1}{s} \times \dfrac {6s}{(s^2 + 9)^2}$

$=\dfrac {6}{(s^2 +9)^2}$

By the shifting property,

$L\Big [ e^{-at} f(t) \Big ] = \phi (s+a)$

$\therefore L \left [ e^{-3u} \displaystyle \int^t_0 u \sin 3u \ du \right ] = \dfrac {6}{\Big [ (s+3)^2 + 9 \Big]^2} $

$= \dfrac {6}{(s^2 + 6s+ 18)^2}$

ii)Let $f(t) = \dfrac {1-\cos 2t}{t} $

Consider, $L[1-\cos 2t] = L[1] - L[\cos 2t]$

$=\dfrac {1}{s} - \dfrac {s} {s^2 +2^2}$

$\displaystyle \therefore L \left ( \dfrac {1- \cos 2t}{t} \right ) = \int^\infty_s \left ( \dfrac {1}{s}- \dfrac {s}{s^2 + 2^2} \right )ds$

$\displaystyle = \dfrac {1}{2} \int^\infty_s \left (\dfrac {2}{s} - \dfrac {2s}{s^2 + 4} \right )ds $

$= \dfrac {1}{2} \Big[2\log s - \log (s^2 +4)\Big]^{\infty}_s $

$= \dfrac {1}{2} \Big [ \log s^2 - \log (s^2 +4) \Big]^{\infty}_s$

$ = \dfrac {1}{2} \left [ \log \left ( \dfrac {s^2}{s^2 +4} \right ) \right ] $

$= \dfrac {1}{2} \left [ 0-\log \left ( \dfrac {s^2}{s^2 +4} \right ) \right ]$

$\therefore L[f(t)] = \dfrac {1}{2}\log \left (\dfrac{s^2 +4}{s^2} \right )$

$\displaystyle now, \ f(0) = \lim_{t\to 0} \dfrac {1-\cos 2t} {t}$

$\displaystyle =\lim_{t\to 0} \dfrac {0+\sin 2t.2}{1}$ (L'Hospital's Rule)

$=0$

Using theorem, 

$L[f'(t)] = - f(0) + sL[f(t)]$

$L\left [ \dfrac {d}{dt} \left ( \dfrac {1-\cos 2t}{t} \right ) \right ]= 0 +s\cdot \dfrac{1}{2} \log \left ( \dfrac {s^2 +4}{s^2} \right )$

$\therefore L \left [ \dfrac {d}{dt} \left ( \dfrac {1-\cos2t}{t} \right ) \right ] = \dfrac {s}{2} \log \left ( \dfrac {s^2 +4}{s^2} \right )$