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If I have a function $f(t)=\sinh(t)\cos(t)$ how would I go about finding the Laplace transform? I tried putting it into the integral defining Laplace transformation: $$ F(s)= \int_0^\infty \mathrm{e}^{-st}\sinh(t)\cos(t)\,\mathrm{d}t $$ But this integral looks very hairy to me. Can $\sinh(t)\cos(t)$ be rewritten as something more manageable perhaps?

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    $\begingroup$ Hint: Look up Convolution Regards $\endgroup$ – Amzoti Sep 15 '13 at 17:58
  • $\begingroup$ The Maple code $$with(inttrans): laplace(sinh(t)*cos(t), t, s); $$ produces $${\frac {{s}^{2}-2}{ \left( \left( s+1 \right) ^{2}+1 \right) \left( \left( s-1 \right) ^{2}+1 \right) }} .$$ $\endgroup$ – user64494 Sep 15 '13 at 18:12
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I would use the exponential formula for $\sinh$, namely $\sinh t = \frac{e^{t}-e^{-t}}{2}$. This turns your integral into $$\frac{1}{2}\int_0^\infty e^{-(s-1)t} \cos t \, dt - \frac{1}{2}\int_0^\infty e^{-(s+1)t} \cos t \, dt$$ Now, if you know what the Laplace transform of the cosine is, you can use that to evaluate these integrals...

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  • $\begingroup$ So the Laplace transform of $cos(t)$ is $\frac{s}{s²+1}$, right? I feel silly asking this, but how do I use this to evaluate the integrals? Thanks. $\endgroup$ – knuterr Sep 15 '13 at 19:28
  • $\begingroup$ Computing the Laplace transform of $\cos t$ is tantamount to integrating $e^{-st} \cos t$. The integrands I have are very similar to that (e.g., to get the the first one, you just have to make the substitution $s \mapsto s-1$). $\endgroup$ – Micah Sep 15 '13 at 19:39
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HINT: $$\sinh t\cos t=\frac{e^t-e^{-t}}2\cdot\frac{e^{it}+e^{-it}}2=\frac{e^{t(1+i)}+e^{t(1-i)}-e^{t(i-1)}-e^{-t(1+i)}}4$$

Now, $$L\{e^{at}\}=\frac1{s-a}$$

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= L [ { (( e^(-t) - e^(t) ))/2 }.{ cost } ]

Taking Constant outside and separating Laplace to different terms,

= (1/2).{ L[ e^(-t).cost ] - L[ e^(t).cost ] }

Separating solving,

L[ e^(-t).cost ],

first solve ----> L[ cost ] = s/ (s^(2) + 1)

Therefore for L[ e^(-t).cost ],

Using shifting property,

L[ e^(-t).cost ] = (s+1)/( (s+1)^2 + 1 ) ...........(1)

similarly,

L[ e^(t).cost ] = (s-1)/( (s-1)^2 + 1 ) ............(2)

Subtracting (1) by (2), will be Answer.

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    $\begingroup$ Welcome to Math.SE. Please use MathJax in your posts to make things easier to read. $\endgroup$ – Ertxiem Apr 6 at 17:32

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