Hey everyone, I am having trouble with this question and was hoping someone can show me how it's done. Find the Laplace transform of the function: f(t) = coshbt (Note: coshx = (e^x + e^(−x))/2)
Follow Math Help Forum on Facebook and Google+
Originally Posted by nak5120 Hey everyone, I am having trouble with this question and was hoping someone can show me how it's done. Find the Laplace transform of the function: f(t) = coshbt (Note: coshx = (e^x + e^(−x))/2) $\displaystyle L{f(t)} = \int_0^{\infty} e^{-st}cosh(bt)~dt$ $\displaystyle = \frac{1}{2} \int_0^{\infty}e^{-st}(e^{bt} + e^{-bt} )~dt$ Are you having problems with this integral? -Dan
I understand the general form of it but I don't get why you would put (1/2) (e^(bt)+e^(-bt) in the equation. Is that supposed to be a solution to cosh(bt) somehow?
Originally Posted by nak5120 Hey everyone, I am having trouble with this question and was hoping someone can show me how it's done. Find the Laplace transform of the function: f(t) = coshbt (Note: coshx = (e^x + e^(−x))/2) You said yourself the definition of cosh(bt). I assumed you would know what that meant when I substituted it in. The solution is the integral that I posted. If you are having problems with the cosh(bt) substitution somehow? -Dan
View Tag Cloud