Question

Calcule a integral por partes

Answer 1

$I=\displaystyle\int\!e^x \cos x\,dx$


Método de integração por partes:

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

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Para avaliar $I_1,$ usaremos o método de integração por partes novamente:

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

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

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


Bons estudos! :-)

Answer 2

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$\displaystyle\sf\int e^x\cdot cos(x)~dx\\\sf fac_{\!\!,}a~u=cos(x)\implies du=-sen(x)~dx\\\sf dv=e^x~dx\implies v=e^x\\\displaystyle\sf\int e^x\cdot cos(x)~dx=e^x\cdot cos(x)-\int e^x~-sen(x)~dx\\\displaystyle\sf\int e^x\cdot cos(x)~dx=e^x\cdot cos(x)+\int e^x\cdot sen(x)~dx\\\sf fac_{\!\!,}a~u_1= sen(x)\implies du_1=cos(x)~dx\\\sf dv_1=e^x\implies v_1=e^x~dx\\\displaystyle\sf\int e^x\cdot sen(x)~dx=e^x\cdot sen(x)-\int e^x\cdot cos(x)~dx$

$\displaystyle\sf\int e^x\cdot cos(x)~dx=e^x\cdot cos(x)+e^x\cdot sen(x)-\int e^x\cdot cos(x)~dx\\\displaystyle\sf\int e^x\cdot cos(x)~dx+\int e^x\cdot cos(x)~dx=e^x[cos(x)+sen(x)]\\\displaystyle\sf2\int e^x\cdot cos(x)~dx=e^x[cos(x)+sen(x)]\\\boxed{\boxed{\boxed{\boxed{\displaystyle\sf\int e^x\cdot cos(x)~dx=\dfrac{1}{2}e^x[cos(x)+sen(x)]+k}}}}$