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Use the definition of Laplace transform to find the transform of f(t)= sinh t. Recall: sinh t=1/2(e^t-e^(-t))
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yes then go for it
\[L(\sinh~at)=\int\limits_{0}^{\infty}e^{-st}.\frac{ e^{at}-e^{-at} }{ 2 }dt\]
can u do it now?
\[L(\sinh~ t)=\int\limits_{0}^{\infty}e^{-st}\frac{ e^t-e^{-t} }{ 2 }\\=\frac{ 1 }{ 2 }[\int\limits_{0}^{\infty}e^{-st}.e^t-\int\limits_{0}^{\infty}e^{-st}.e^{-t}]\\=\frac{ 1 }{ 2 }[L(e^{t})-L(e^{-t})]=\frac{ 1 }{ 2 }[\frac{ 1 }{ s-1 }-\frac{ 1 }{ s+1 }]\] i believe u can do the rest :) @keynote
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