Question

Qual a integral de e^-2x × senx dx​

Answer 1

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador https://brainly.com.br/tarefa/39981882

                                                       

Integral por partes

$\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int\! u\cdot dv=u\!\cdot v\!-\!\int\! v\cdot\! du}}}}$

$\underline{\rm para~resolver~esta~quest\tilde ao~vamos~usar~o~macete~do~ILATE}\\\Large\boxed{\begin{array}{l}\sf I\longrightarrow inversa~trigonom\acute etrica\\\sf L\!\longrightarrow\ell ogaritmo\\\sf A\!\longrightarrow aritm\acute etica\\\sf T\!\longrightarrow trigonom\acute etrica\\\sf E\!\longrightarrow exponencial\\\tt vale~lembrar~que~isso~se~trata~de\\\tt de~um~crit\acute erio~para~escolha~do~''u''\\\tt visando~facilitar~o~c\acute alculo~da~integral\end{array}}$

$\displaystyle\sf\int e^{-2x}\cdot sen\,x~dx\\\underline{\rm fac_{\!\!,}a}\\\sf u= sen\,x\implies du=cos\,x~dx\\\sf dv=e^{-2x}\implies v=-\dfrac{1}{2}e^{-2x}\\\displaystyle\sf\int e^{-2x}\cdot sen\,x~dx=-\dfrac{1}{2}e^{-2x}\cdot sen\,x-\int-\dfrac{1}{2}e^{-2x}\cdot cos\,x~dx\\\displaystyle\sf\int e^{-2x}\cdot sen\,x~dx=-\dfrac{1}{2}e^{-2x}\cdot sen\,x+\dfrac{1}{2}\int e^{-2x}\cdot cos\,x~dx\\\underline{\rm fac_{\!\!,}a}\\\sf u_1=cos\,x\implies du_1=-sen\,x~dx\\\sf dv_1=e^{-2x}\to v_1=-\dfrac{1}{2}e^{-2x}$

$\displaystyle\sf\int e^{-2x}\cdot cos\,x~dx=-\dfrac{1}{2}e^{-2x}\cdot cos\,x-\int-\dfrac{1}{2}e^{-2x}\cdot -sen\,x~dx\\\displaystyle\sf\int e^{-2x}\cdot cos\,x~dx=-\dfrac{1}{2}e^{-2x}\cdot cos\,x-\dfrac{1}{2}\int e^{-2x}\cdot sen\,x~dx.$

$\displaystyle\sf\int e^{-2x}\cdot sen\,x~dx=-\dfrac{1}{2}e^{-2x}\cdot sen\,x+\dfrac{1}{2}\bigg[-\dfrac{1}{2}e^{-2x}\cdot cos\,x-\dfrac{1}{2}\int e^{-2x}\cdot sen\,x~dx\bigg]\\\displaystyle\sf\int e^{-2x}\cdot sen\,x~dx=-\dfrac{1}{2}e^{-2x}\cdot sen\,x-\dfrac{1}{4}e^{-2x}\cdot cos\,x-\dfrac{1}{4}\int e^{-2x}\cdot sen\,x~dx\\\displaystyle\sf\int e^{-2x}\cdot sen\,x~dx+\dfrac{1}{4}\int e^{-2x}\cdot sen\,x~dx=-\dfrac{1}{2}e^{-2x}\bigg[sen\,x+\dfrac{1}{2}\cdot cos\,x\bigg]$

$\blue{\small\boxed{\begin{array}{l}\displaystyle\sf\dfrac{5}{4}\int e^{-2x}\cdot sen\,x~dx=-\dfrac{1}{2}e^{-2x}\bigg[sen\,x+\dfrac{1}{2}\cdot cos\,x\bigg]\\\displaystyle\sf\int e^{-2x}\cdot sen\,x~dx=\dfrac{4}{5}\cdot-\dfrac{1}{2}e^{-2x}\bigg[sen\,x+\dfrac{1}{2}\cdot cos\,x\bigg]+k\\\displaystyle\sf\int e^{-2x}\cdot sen\,x~dx=-\dfrac{2}{5}e^{-2x}\bigg[sen\,x+\dfrac{1}{2}\cdot cos\,x\bigg]+k\end{array}}}$