Find the Laplace transform of $(t-\pi/2)\sin(t-\pi/2)$ using the time shift

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What is the Laplace transform of $(t-\pi/2)\sin(t-\pi/2)$?

I used the relationship $\mathcal{L}((t-a)f(t-a))=e^{-as}F(s)$

Hence I get $\dfrac{2e^{-(\pi/2)s}}{s^2+1}$. Would this be correct?