Question

Qual o valor da integral indefinida $\int\limits { e^{-x} } \,cos2x dx$ veja opções na foto

Answer 1

Integral indefinida:

$I_{0}=\int{e^{-x}\cos 2x\,dx}$


$\bullet\;\;$ Utilizaremos o método de integração por partes:

$\begin{array}{ll} u_{0}=e^{-x}\;\;&\;\;du_{0}=-e^{-x}\,dx\\ \\ dv_{0}=\cos 2x\,dx\;\;&\;\;v_{0}=\dfrac{1}{2}\mathrm{\,sen\,}2x \end{array}\\ \\ \\ \int{u_{0}\,dv_{0}}=u_{0}\,v_{0}-\int{v_{0}\,du_{0}}\\ \\ \\ \int{e^{-x}\cos 2x\,dx}=\dfrac{1}{2}\,e^{-x}\mathrm{\,sen\,}2x-\int{\dfrac{1}{2}\mathrm{\,sen\,}2x\cdot (-e^{-x})\,dx}\\ \\ \\ \int{e^{-x}\cos 2x\,dx}=\dfrac{1}{2}\,e^{-x}\mathrm{\,sen\,}2x+\dfrac{1}{2}\int{e^{-x}\mathrm{\,sen\,}2x\,dx}\\ \\ \\ I_{0}=\dfrac{1}{2}\,e^{-x}\mathrm{\,sen\,}2x+\dfrac{1}{2}\,I_{1}\;\;\;\;\;(i)$

onde 

$I_{1}=\int{e^{-x}\mathrm{\,sen\,}2x\,dx}$


$\bullet\;\;$ Novamente, por partes, temos

$\begin{array}{ll} u_{1}=e^{-x}\;\;&\;\;du_{1}=-e^{-x}\,dx\\ \\ dv_{1}=\mathrm{sen\,}2x\,dx\;\;&\;\;v_{1}=-\dfrac{1}{2}\cos 2x \end{array}\\ \\ \\ \int{u_{1}\,dv_{1}}=u_{1}\,v_{1}-\int{v_{1}\,du_{1}}\\ \\ \\ \int{e^{-x}\mathrm{\,sen\,}2x\,dx}=-\dfrac{1}{2}\,e^{-x}\cos 2x-\int{-\dfrac{1}{2}\cos 2x\cdot (-e^{-x})\,dx}\\ \\ \\ \int{e^{-x}\mathrm{\,sen\,}2x\,dx}=-\dfrac{1}{2}\,e^{-x}\cos 2x-\dfrac{1}{2}\int{e^{-x}\cos 2x\,dx}\\ \\ \\ I_{1}=-\dfrac{1}{2}\,e^{-x}\cos 2x-\dfrac{1}{2}\,I_{0}\;\;\;\;\;(ii)$


Substituindo $(ii)$ em $(i),$ temos

$I_{0}=\dfrac{1}{2}\,e^{-x}\mathrm{\,sen\,}2x+\dfrac{1}{2}\cdot\left[-\dfrac{1}{2}\,e^{-x}\cos 2x-\dfrac{1}{2}\,I_{0} \right ]\\ \\ \\ I_{0}=\dfrac{1}{2}\,e^{-x}\mathrm{\,sen\,}2x-\dfrac{1}{4}\,e^{-x}\cos 2x-\dfrac{1}{4}\,I_{0} \\ \\ \\ I_{0}+\dfrac{1}{4}\,I_{0}=\dfrac{1}{2}\,e^{-x}\mathrm{\,sen\,}2x-\dfrac{1}{4}\,e^{-x}\cos 2x\\ \\ \\ \left(1+\dfrac{1}{4} \right)\,I_{0}=\dfrac{1}{2}\,e^{-x}\mathrm{\,sen\,}2x-\dfrac{1}{4}\,e^{-x}\cos 2x\\ \\ \\ \dfrac{4+1}{4}\,I_{0}=\dfrac{1}{2}\,e^{-x}\mathrm{\,sen\,}2x-\dfrac{1}{4}\,e^{-x}\cos 2x\\ \\ \\ \dfrac{5}{4}\,I_{0}=\dfrac{1}{2}\,e^{-x}\mathrm{\,sen\,}2x-\dfrac{1}{4}\,e^{-x}\cos 2x$


Multiplicando os dois lados por $\dfrac{4}{5},$ chegamos a

$I_{0}=\dfrac{2}{5}\,e^{-x}\mathrm{\,sen\,}2x-\dfrac{1}{5}\,e^{-x}\cos 2x+C\\ \\ \\ I_{0}=\dfrac{e^{-x}}{5}\left(2\,\mathrm{\,sen\,}2x-\cos 2x \right )+C\\ \\ \\ \boxed{ \int{e^{-x}\cos 2x\,dx}=\dfrac{e^{-x}}{5}\left(2\mathrm{\,sen\,}2x-\cos 2x \right )+C }$