Here's the question you clicked on:
UnkleRhaukus
\[ \mathcal L ^{-1}\left\{\frac{e^{-\pi p}}{p^2+1}\right\}\]
\[ \begin{align*} \mathcal L ^{-1}\left\{\frac{e^{-\pi p}}{p^2+1}\right\} &=h(t-\pi)\times\mathcal L^{-1} \left\{\frac1{p^2+1}\right\}_{t\rightarrow t-\pi}\\ &=h(t-\pi)\times\sin(t)\Big|_{t\rightarrow t-\pi}\\ &=h(t-\pi)\times\sin(t-\pi)\\ &=-h(t-\pi)\sin(t)\\ \end{align*}\]
have i got this one right/?
This appears correct to me. :)
thanks!, would you more likely leave the answer in the final form of \[=h(t−π)\sin(t−π)\]or \[=−h(t−π)\sin(t)\]
I would probably use -h(t-pi) sin(t) just because I prefer to see the simplest forms of functions..