$$L[sinu] = \frac{1}{s^2 + 1}$$

$L\bigg(\frac{sinu}{u}\bigg) = \int\limits_{s}^{\infty}\frac{1}{(s^2 + 1)}ds$

Division property

$L\bigg(\frac{sinu}{u}\bigg) = \big[tan^{-1}s\big]_{s}^{\infty}$

$L\bigg(\frac{sinu}{u}\bigg) = \frac{\pi}{2} - tan^{-1}s$

$L\bigg(\frac{sinu}{u}\bigg) = cot^{-1}s$

Now $L\bigg[\int\limits_{0}^{t}\frac{sinu}{u}du\bigg] = \frac{1}{s}cot^{-1}s$

By integration property

Now $L\bigg[e^{-t}\int\limits_{0}^{t}\frac{sinu}{u}du\bigg] = \frac{1}{s + 1}cot^{-1}(s + 1)$

By first shifting