$\mathrm{i)~e^{-t}\displaystyle\int_0^t \dfrac{sin~u}{u}~du}$ Now, $\mathrm{L\Big[sin~u\Big]=\dfrac{1}{s^2+1}}$ $\mathrm{L\Big[\dfrac{sin~u}{u}\Big]=\displaystyle\int_s^\infty \phi(s)~ds=\displaystyle\int_s^\infty\dfrac{ds}{s^2+1}}$ $\mathrm{=\Big[ tan^{-1}~s \Big]_s^\infty=\dfrac{\pi}{2}-tan^{-1}~s=cot^{-1}~s}$ $\mathrm{\therefore L\bigg[\displaystyle\int_0^t\dfrac{sin~u}{u}~du\bigg]=\dfrac{1}{s}\cdot\phi(s)=\dfrac{1}{s}\cdot cot^{-1}~s}$ $\mathrm{\therefore L\bigg[e^{-t}\displaystyle\int_0^t\dfrac{sin~u}{u}~du\bigg]=\dfrac{1}{(s+1)}cot^{-1}(s+1)}$   $\mathrm{ii)~t\sqrt{1+sin~t}}$ We have, $\mathrm{\sqrt{1+sin~t}=\sqrt{\bigg[sin^2(t/2)+cos^2(t/2)+2~sin(t/2)cos(t/2)\bigg]}}$ $\mathrm{\therefore \sqrt{1+sin~t}=\sqrt{\big[sin(t/2)+cos(t/2)\big]^2}~=sin(t/2)+cos(t/2)}$ $\mathrm{\therefore L \Big[\sqrt{1+sin~t}\Big]= L \Big[sin(t/2)+cos(t/2)\Big]}$ $\mathrm{=\dfrac{1/2}{s^2+(1/2)^2}+\dfrac{s}{s^2+(1/2)^2}=\dfrac{1}{2}\cdot \dfrac{4}{(4s^2+1)}+\dfrac{4s}{(4s^2+1)}}$ $\mathrm{=\dfrac{4s+2}{(4s^2+1)}=\dfrac{2(2s+1)}{(4s^2+1)}}$ $\mathrm{\therefore L \Big[t\sqrt{1+sin~t}\Big]= -\dfrac{d}{ds}\bigg[\dfrac{2(2s+1)}{(4s^2+1)}\bigg]=-2\bigg[\dfrac{(4s^2+1)2-(2s+1)8s}{(4s^2+1)^2}\bigg]}$ $\mathrm{=-2\dfrac{\bigg(-8s^2-8s+2\bigg)}{(4s^2+1)^2}=4\dfrac{\big(4s^2+4s-1\big)}{(4s^2+1)^2}}$ $\mathrm{\therefore L \Big[t\sqrt{1+sin~t}\Big]=4\dfrac{\bigg(4s^2+4s-1\bigg)}{(4s^2+1)^2} }$