Question

integral de e^-x sen(x) dx?
Por partes.

Answer 1

$I=\displaystyle\int e^{-x}\,\mathrm{sen\,}x\,dx$


Método de integração por partes:

$\begin{array}{lcl} u=e^{-x}&\Rightarrow&du=-\,e^{-x}\,dx\\\\ dv=\mathrm{sen\,}x\,dx&\Leftarrow&v=-\cos x \end{array}\\\\\\\\ \displaystyle\int u\,dv=uv-\int v\,du\\\\\\ \int e^{-x}\,\mathrm{sen\,}x\,dx=-\,e^{-x}\cos x-\int (-\cos\,x)\cdot (-\,e^{-x})\,dx\\\\\\ I=-\,e^{-x}\cos x-\underbrace{\int e^{-x}\cos x\,dx}_{I_1}~~~~~~\mathbf{(i)}$

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Calculando $I_1:$

$I_1=\displaystyle\int e^{-x}\cos x\,dx$


Método de integração por partes:

$\begin{array}{lcl} u=e^{-x}&\Rightarrow&du=-\,e^{-x}\,dx\\\\ dv=\cos x\,dx&\Leftarrow&v=\mathrm{sen\,} x \end{array}\\\\\\\\ \displaystyle\int u\,dv=uv-\int v\,du\\\\\\ \int e^{-x}\cos x\,dx=e^{-x}\,\mathrm{sen\,}x-\int \mathrm{sen\,}x\cdot (-\,e^{-x})\,dx\\\\\\ \int e^{-x}\cos x\,dx=e^{-x}\,\mathrm{sen\,}x+\int e^{-x}\,\mathrm{sen\,}x\,dx\\\\\\ I_1=e^{-x}\,\mathrm{sen\,}x+I~~~~~~\mathbf{(ii)}$

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Substituindo $\mathbf{(ii)}$ em $\mathbf{(i)},$ temos

$I=-\,e^{-x}\cos x-(e^{-x}\,\mathrm{sen\,}x+I)\\\\ I=-\,e^{-x}\cos x-\,e^{-x}\,\mathrm{sen\,}x-I\\\\ I+I=-\,e^{-x}\cos x-\,e^{-x}\,\mathrm{sen\,}x\\\\ 2I=-\,e^{-x}\cos x-\,e^{-x}\,\mathrm{sen\,}x\\\\ I=-\,\dfrac{1}{2}e^{-x}\cos x-\dfrac{1}{2}\,e^{-x}\,\mathrm{sen\,}x+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\displaystyle\int e^{-x}\,\mathrm{sen\,}x\,dx=-\,\dfrac{1}{2}e^{-x}\cos x-\dfrac{1}{2}\,e^{-x}\,\mathrm{sen\,}x+C \end{array}}$


Bons estudos! :-)