Inverse Laplace Transform without using equations I'm stuck with a question

The inverse Laplace transform of ${\displaystyle \frac{1}{s^2+1}}$ is ${\displaystyle \sin t}$. Therefore we have
${\mathcal{L}}^{-1}\left\{\frac{1}{(s+1{)}^2+1}\right\}=e^{-t}\sin \left(t\right)=\frac{1}{2i}(e^{(-1+i)t}-e^{(-1-i)t})$
Therefore
${\mathcal{L}}^{-1}\left\{\frac{1}{s}\frac{1}{(s+1{)}^2+1}\right\}={\int }_0^t{e}^{-u}\sin \left(u\right)du=\frac{1}{2i}(\frac{e^{(-1+i)t}-1}{-1+i}-\frac{e^{(-1-i)t}-1}{-1-i})$
${\displaystyle =\frac{1}{2}-\frac{1}{2}{e}^{-t}(\sin \left(t\right)+\cos \left(t\right))}$
The last step is to use the shift theorem ${\displaystyle {\mathcal{L}}^{-1}\left\{e^{-as}F\left(s\right)\right\}=f(t-a)\mathrm{\Theta }(t-a)}$ and we get
${\mathcal{L}}^{-1}\left\{{\mathrm{e}}^{-\mathrm{\pi s}}\frac{1}{\mathrm{s}}\frac{1}{(\mathrm{s}+1{)}^2+1}\right\}=[\frac{1}{2}-\frac{1}{2}{\mathrm{e}}^{-(\mathrm{t}-\mathrm{\pi })}(\sin (\mathrm{t}-\mathrm{\pi })+\cos (\mathrm{t}-\mathrm{\pi })\left)\right]\mathrm{\Theta }(\mathrm{t}-\mathrm{\pi })$
$=(\frac{1}{2}+\frac{1}{2}{\mathrm{e}}^{\mathrm{\pi }-\mathrm{t}}(\cos \left(\mathrm{t}\right)+\sin \left(\mathrm{t}\right)\left)\right)\mathrm{\Theta }(\mathrm{t}-\mathrm{\pi })$