Answer:
L[f(t)=s/(1+s^2)]
Step-by-step explanation:
The Laplace Transform is given by the integral:
[tex]L[f(t)]=\int_0^\infty e^{-st}\ f(t)dt[/tex]
by replacing f(t)=cost we get
[tex]\int_0^{\infty} e^{-st}costdt=[e^{-s(\infty)}sin(\infty)-1sin(0)]-s\int_{0}^{\infty}e^{-st}sintdt\\\\=0+s[-e^{-s(\infty)}cos(\infty)+e^{s(0)}cost(0)-s\int_0^{\infty}e^{-st}costdt]\\\\=0+s[0+1-s\int_0^{\infty}e^{-st}costdt]=s-s^2\int_0^{\infty}e^{-st}costdt\\\\(1+s^2)\int_0^{\infty}e^{-st}costdt=s\\\\\int_0^{\infty}e^{-st}costdt=\frac{s}{1+s^2}[/tex]
hope this helps!!
