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# Laplace de cos(at)

Alguém pode me dar a resolução para a transformada de Laplace de cos(at), sendo "a" uma constante pela definição?

## 4 resposta(s) - Contém resposta de Especialista

RD Resoluções

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\begin{align} & L\{\cos (at)\}=\underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{-st}}\cos (at)dt \\ & Logo\text{ }c=ib: \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{ct}}{{e}^{-st}}dt=\underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{-t(s-c)}}=\frac{{{e}^{-t(s-c)}}}{-(s-c)}|_{0}^{\infty }=\frac{1}{s-c}\text{ }\!\!~\!\!\text{ para }\!\!~\!\!\text{ }s-c>0 \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{(ib)t}}{{e}^{-st}}dt=\frac{1}{s-ib} \\ & Usando\text{ }Euler: \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,(\cos bt+i\operatorname{s}\text{e}nbt){{e}^{-st}}dt=\frac{1}{s-ib}=\frac{s+ib}{{{s}^{2}}+{{b}^{2}}} \\ & \text{Igualando as partes reais:} \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,(\cos bt){{e}^{-st}}dt=\frac{s}{{{s}^{2}}+{{b}^{2}}} \\ & \int {{e}^{at}}\cos btdt=\cos bt\int {{e}^{at}}dt-\int \left( \frac{\cos bt}{dt}\int {{e}^{at}}dt \right)dt=\frac{{{e}^{at}}\cos bt}{a}+\frac{b}{a}\int {{e}^{at}}\sin btdt \\ & Logo: \\ & \int {{e}^{at}}\cos btdt=\frac{{{e}^{at}}\cos bt}{a}+\frac{b}{a}\int {{e}^{at}}\operatorname{s}enbtdt \\ & I=\int {{e}^{at}}\cos btdt=\frac{{{e}^{at}}\cos bt}{a}+\frac{b}{a}\left( \frac{{{e}^{at}}\operatorname{s}enbt}{a}-\frac{b}{a}\int {{e}^{at}}\cos btdt \right) \\ & \;\;I=\frac{{{e}^{at}}\cos bt}{a}+\frac{b}{{{a}^{2}}}{{e}^{at}}\operatorname{s}enbt-\frac{{{b}^{2}}}{{{a}^{2}}}I\to \;\;\int {{e}^{at}}\cos btdt=I=\frac{{{e}^{at}}(a\cos bt+b\operatorname{s}enbt)}{{{a}^{2}}+{{b}^{2}}} \\ & a=-s,\int {{e}^{-st}}\cos btdt=I=\frac{{{e}^{-st}}(-s\cos bt+b\operatorname{s}enbt)}{{{s}^{2}}+{{b}^{2}}} \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,(\cos bt){{e}^{-st}}dt=\frac{{{e}^{-st}}(-s\cos bt+b\operatorname{s}enbt)}{{{s}^{2}}+{{b}^{2}}}|_{0}^{\infty }=\frac{s}{{{s}^{2}}+{{b}^{2}}} \\ & \\ & \\ \end{align}

\begin{align} & L\{\cos (at)\}=\underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{-st}}\cos (at)dt \\ & Logo\text{ }c=ib: \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{ct}}{{e}^{-st}}dt=\underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{-t(s-c)}}=\frac{{{e}^{-t(s-c)}}}{-(s-c)}|_{0}^{\infty }=\frac{1}{s-c}\text{ }\!\!~\!\!\text{ para }\!\!~\!\!\text{ }s-c>0 \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,{{e}^{(ib)t}}{{e}^{-st}}dt=\frac{1}{s-ib} \\ & Usando\text{ }Euler: \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,(\cos bt+i\operatorname{s}\text{e}nbt){{e}^{-st}}dt=\frac{1}{s-ib}=\frac{s+ib}{{{s}^{2}}+{{b}^{2}}} \\ & \text{Igualando as partes reais:} \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,(\cos bt){{e}^{-st}}dt=\frac{s}{{{s}^{2}}+{{b}^{2}}} \\ & \int {{e}^{at}}\cos btdt=\cos bt\int {{e}^{at}}dt-\int \left( \frac{\cos bt}{dt}\int {{e}^{at}}dt \right)dt=\frac{{{e}^{at}}\cos bt}{a}+\frac{b}{a}\int {{e}^{at}}\sin btdt \\ & Logo: \\ & \int {{e}^{at}}\cos btdt=\frac{{{e}^{at}}\cos bt}{a}+\frac{b}{a}\int {{e}^{at}}\operatorname{s}enbtdt \\ & I=\int {{e}^{at}}\cos btdt=\frac{{{e}^{at}}\cos bt}{a}+\frac{b}{a}\left( \frac{{{e}^{at}}\operatorname{s}enbt}{a}-\frac{b}{a}\int {{e}^{at}}\cos btdt \right) \\ & \;\;I=\frac{{{e}^{at}}\cos bt}{a}+\frac{b}{{{a}^{2}}}{{e}^{at}}\operatorname{s}enbt-\frac{{{b}^{2}}}{{{a}^{2}}}I\to \;\;\int {{e}^{at}}\cos btdt=I=\frac{{{e}^{at}}(a\cos bt+b\operatorname{s}enbt)}{{{a}^{2}}+{{b}^{2}}} \\ & a=-s,\int {{e}^{-st}}\cos btdt=I=\frac{{{e}^{-st}}(-s\cos bt+b\operatorname{s}enbt)}{{{s}^{2}}+{{b}^{2}}} \\ & \underset{0}{\overset{\infty }{\mathop \int }}\,(\cos bt){{e}^{-st}}dt=\frac{{{e}^{-st}}(-s\cos bt+b\operatorname{s}enbt)}{{{s}^{2}}+{{b}^{2}}}|_{0}^{\infty }=\frac{s}{{{s}^{2}}+{{b}^{2}}} \\ & \\ & \\ \end{align}

Luiz Francisco Batista Sampaio

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Boa tarde,

Coloquei a resposta no link: https://www.passeidireto.com/arquivo/3381239/lcosat

Espero que ajude! Bons estudos!

Marcos Souza

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