# Solve f(t)=e^t cos t

Solve $f\left(t\right)={e}^{t}\mathrm{cos}t$
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averes8
Step 1
We have to form Laplace transform integral first.
Step 2
We have $f\left(t\right)={e}^{t}\mathrm{cos}t$
We know that,
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$⇒F\left(s\right)={\int }_{0}^{\mathrm{\infty }}\mathrm{cos}\left(t\right){e}^{\left(1-s\right)t}dt$
by integrath by pcuts
$⇒F\left(s\right)={\left[\frac{\mathrm{cos}\left(t\right){e}^{\left(1-s\right)t}}{1-s}\right]}_{0}^{\mathrm{\infty }}+\int \left(\frac{{e}^{\left(1-s\right)t}}{1-s}\right)\mathrm{sin}tdt$
$⇒F\left(s\right)=-\frac{1}{1-s}+\frac{1}{1-s}{\left[\frac{{e}^{\left(1-s\right)t}\mathrm{sin}t}{1-s}\right]}_{0}^{\mathrm{\infty }}-\int \left(\frac{{e}^{\left(1-s\right)t}}{1-s}\right)\mathrm{cos}tdt$
$⇒F\left(s\right)=-\frac{1}{1-s}-\frac{1}{\left(1-s{\right)}^{2}}$
$⇒F\left(s\right)\left[1+\frac{1}{\left(1-s{\right)}^{2}}\right]=-\frac{1}{\left(1-s\right)}$
$⇒F\left(s\right)=\frac{-\left(1-s{\right)}^{2}}{\left(1-s\right)\left[\left(1-s{\right)}^{2}+1\right]}$
$⇒F\left(s\right)=\frac{s-1}{\left[\left(s-1{\right)}^{2}+1\right]}$
Answer: $F\left(s\right)=\frac{s-1}{\left[\left(s-1{\right)}^{2}+1\right]}$