Use any method to find the laplace transform of coshbt
Looking to get help with this example for my exam review
The laplace transformations of $coshbt$ is the following $$\int_0^\infty cosh({bt})e^{-st} dt$$ $$= \int_0^\infty \frac{(e^bt + e^{-bt})e^{-st}}{2} dt$$ $$= \frac{1}{2} \int_0^\infty e^{-st + bt} + e^{-st - bt} dt $$ $$= \frac{1}{2} \int_0^\infty e^{(-s + b)t} + e^{(-s - b)t} dt$$ $$=\frac{1}{2} \begin{bmatrix} \frac {e^{(-s+ b)t}}{-s+b} + \frac{e^{-st - bt}}{-s-b} \end{bmatrix}^\infty_0$$
You can probably take it from here.
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The laplace transformations of $coshbt$ is the following $$\int_0^\infty cosh({bt})e^{-st} dt$$ $$= \int_0^\infty \frac{(e^bt + e^{-bt})e^{-st}}{2} dt$$ $$= \frac{1}{2} \int_0^\infty e^{-st + bt} + e^{-st - bt} dt $$ $$= \frac{1}{2} \int_0^\infty e^{(-s + b)t} + e^{(-s - b)t} dt$$ $$=\frac{1}{2} \begin{bmatrix} \frac {e^{(-s+ b)t}}{-s+b} + \frac{e^{-st - bt}}{-s-b} \end{bmatrix}^\infty_0$$
You can probably take it from here.