Respuesta :
Answer:
[tex]F(s) = \frac{s(e^{\pi s}+1)}{s^2 +1}[/tex]
Step-by-step explanation:
Using the formula for Laplace the transformations if [tex]F(s)[/tex] is the converted function then
[tex]F(s) = \int\limits_{0}^{\infty} e^{-st} \cos(t) dt = \int\limits_{0}^{\pi} e^{-st} \cos(t) dt[/tex]
To solve that integral you need to use integration by parts, when you do integration by parts you get that
[tex]F(s) = \frac{s(e^{\pi s}+1)}{s^2 +1}[/tex].
Answer:
The laplace transform is [tex] F(s) = \frac{s(1+e^{-s\pi})}{s^2+1}[/tex]
Step-by-step explanation:
Let us asume that f(t) =0 for t<0. So, by definition, the laplace transform is given by:
[tex]I = \int_{0}^\pi e^{-st}\cos(t) dt[/tex]
To solve this integral, we will use integration by parts. Let u= cos(t) and dv = [tex]e^{-st}[/tex], so v=[tex]\frac{-e^{st}}{s}[/tex] and du = -sin(t), then, in one step of the integration we have that
[tex]I = \left.\frac{-\cos(t) e^{-st}}{s}\right|_{0}^\pi- \int_{0}^\pi \frac{\sin(t) e^{-st}}{s} dt[/tex]
Let [tex] I_2 = \int_{0}^\pi \frac{\sin(t) e^{-st}}{s} dt[/tex]. We will integrate I_2 again by parts. Choose u = sin(t) and dv = [tex]\frac{e^{-st}}{s}[/tex]. So
[tex] I_2 = \left.\frac{-\sin(t) e^{-st}}{s^2}\right|_{0}^\pi + \int_{0}^\pi \frac{\cos(t) e^{-st}}{s^2}dt [/tex]
Therefore,
[tex]I = \left.\frac{-\cos(t) e^{-st}}{s}\right|_{0}^\pi - (\left.\frac{-\sin(t) e^{-st}}{s^2}\right|_{0}^\pi - \frac{1}{s^2} I[/tex]
which is an equation for the variabl I. Solving for I we have that
[tex]I(\frac{s^2+1}{s^2}) =\left.\frac{-\cos(t) e^{-st}}{s}\right|_{0}^\pi+\left.\frac{\sin(t) e^{-st}}{s^2}\right|_{0}^\pi[/tex]
Then,
[tex]I = \left.\frac{-s\cos(t) e^{-st}}{s^2+1}\right|_{0}^\pi+\left.\frac{\sin(t) e^{-st}}{s^2+1}\right|_{0}^\pi[/tex].
Note that since the sine function is 0 at 0 and pi, we must only care on the first term. Then
[tex]I = \left.\frac{-s\cos(t) e^{-st}}{s^2+1}\right|_{0}^\pi = \frac{s}{s^2+1}(1-(-1)e^{-s\pi}} = \frac{s(1+e^{-s\pi})}{s^2+1}[/tex]
