To find Laplace Transform of f (t) = $\int_0^t u\:e^{-3u} cos^2{2u}\:\mathrm{d}u$

We know that $cos^22u ={{1+cos4u}\over2}$

 $\therefore\: L[cos^22u]={1\over2}L[1+cos4u]={1\over2}L[1]+{1\over2}{L[cos4u]}$

 $\therefore\: L[cos^22u]$ =${1\over2}\cdot{1\over s}+{1\over2}\cdot{s\over{s^2+4^2}}$

Now,

$L[{e^{-3u}}\:cos^22u]={1\over2}\cdot{1\over{s+3}}+{1\over2}{{s+3}\over{(s+3)^2+4^2}}={1\over2}({{1\over s+3}+{s+3\over{(({s+3})^2)+4^2}})}$

                     =${1\over2}({{1\over s+3}+{s+3\over{s^2+6s+25}}})$

$\therefore\:L[u\:{e^{-3u}}\:cos^22u]={-d\over ds}{1\over2}{({1\over s+3}+{s+3\over({s+3})^2+4^2})}$

                            =${1\over2}({1\over{(s+3})^2}-[{(s^2+6s+25)-(s+3)(2s+6)\over(s^2+6s+25)^2}])={1\over2}({{1\over(s+3)^2}+{{s^2+6s-7}\over(s^2+6s+25)^2}})=\phi(s)...(say)$

$\therefore\:L[\int_0^t u\:e^{-3u} cos^2{2u}\:\mathrm{d}u]$ = ${1\over s}\:{\phi(s)}$ = ${1\over s}\cdot{{1\over2}({{1\over(s+3)^2}+{{s^2+6s-7}\over(s^2+6s+25)^2}})}$

$\therefore\:L[\int_0^t u\:e^{-3u} cos^2{2u}\:\mathrm{d}u]$ = ${{1\over2s}({{1\over(s+3)^2}+{{s^2+6s-7}\over(s^2+6s+25)^2}})}$