Let f(t) be a function defined for all positive values of t. The Laplace Transform of f(t) is defined by the following integral, if the improper integral exists.F(s) = ∞e−st f(t) dt0 Laplace Transforms are used to solve differential equations. Find the Laplace Transform of the function.f(t) = cos at

Answer:We have laplace transform of any function is defined as [tex]L[{f(t)}]=\int_{0}^{\infty }f(t)e^{-st}dt[/tex]thus for [tex]f(t)=cos(at)[/tex] [tex]L[cos(at)]=\int_{0}^{\infty }cos(at)e^{-st}dt[/tex]Now integrating the given expression using the integration by parts principle we have[tex]\int cos(at)e^{-st}dt=cos(at)\int e^{-st}dt-\int \frac{dcos(at)}{dt}\int (e^{-st}dt)dt\\\\=cos(at)\frac{e^{-st}}{-s}-(\int -asin(at)\frac{e^{-st}}{-s})\\\\-\frac{cos(at)e^{-st}}{s}-\frac{a}{s}\int sin(at)e^{-st}dt\\\\=-\frac{cos(at)e^{-st}}{s}-\frac{a}{s}(sin(at)\int e^{-st}dt-\int \frac{dsin(at)}{dt}\int (e^{-st}dt)dt\\\\=-\frac{cos(at)e^{-st}}{s}-\frac{a}{s}[(\frac{sin(at)e^{-st}}{-s})-\int acos(at)\frac{e^{-st}}{-s}[/tex]Note that we get the first integral back in the RHS expression Thus solving we have[tex]I=-\frac{cos(at)e^{-st}}{s}+\frac{asin(at)e^{-st}}{s^{2}}+\frac{a}{s}I[/tex]Solving for I we get[tex]I=\frac{e^{-st}(-scos(at)+asin(at))}{s^{2}+a^{2}}[/tex]now applying the limits we have [tex]I=\frac{e^{-st}(-scos(at)+asin(at))}{s^{2}+a^{2}}|_{0}^{\infty }\\\\I=\frac{s}{s^{2}+a^{2}}(\because \lim_{t\rightarrow \infty}e^{-st}=0)[/tex]