Question

Resolva a integra ∫ e^3xcos4xdx pelo método da integração por partes

Answer 1

Faça

$u=cos(4x)dx~~~~~~~~~\longrightarrow~~~~~~~~~du=-4sen(4x)dx\\\\v=\frac{1}{3}e^{3x}~~~~~~~~~~~~~~~~\longleftarrow~~~~~~~~~dv=e^{3x}dx$

Pela fórmula da integração por partes:

$\displaystyle\int udv=uv-\int vdu\\\\\\\int e^{3x}cos(4x)dx=\dfrac{e^{3x}}{3}cos(4x)-\int\dfrac{1}{3}e^{3x}(-4)sen(4x)dx\\\\\\\int e^{3x}cos(4x)dx=\dfrac{e^{3x}}{3}cos(4x)+\dfrac{4}{3}\int e^{3x}sen(4x)dx$

Chamando a integral que queremos de I:

$\boxed{\boxed{I=\dfrac{e^{3x}}{3}cos(4x)+\dfrac{4}{3}\int e^{3x}sen(4x)dx}}~~(i)$
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Agora, vamos achar uma forma para a integral encontrada, também usando integração por partes

$\displaystyle\int e^{3x}sen(4x)dx$

Fazendo as substituições

$a=sen(4x)~~~~~~~\longrightarrow~~~~~~~da=4cos(4x)dx\\\\b=\frac{1}{3}e^{3x}~~~~~~~~~~~\longleftarrow~~~~~~~db=e^{3x}dx$

Pela integração por partes:

$\displaystyle\int e^{3x}sen(4x)dx=\dfrac{e^{3x}}{3}sen(4x)-\int\dfrac{1}{3}e^{3x}\cdot4cos(4x)dx\\\\\\\int e^{3x}sen(4x)dx=\dfrac{e^{3x}}{3}sen(4x)-\dfrac{4}{3}\int e^{4x}cos(4x)dx\\\\\\\boxed{\boxed{\int e^{3x}sen(4x)dx=\dfrac{e^{3x}}{3}sen(4x)-\dfrac{4}{3}I}}~~(ii)$

Substituindo (ii) em (i)

$I=\displaystyle\dfrac{e^{3x}}{3}cos(4x)+\dfrac{4}{3}\int e^{3x}sen(4x)dx\\\\\\I=\dfrac{e^{3x}}{3}cos(4x)+\dfrac{4}{3}\left(\dfrac{e^{3x}}{3}sen(4x)-\dfrac{4}{3}I\right)+C\\\\\\I=\dfrac{e^{3x}}{3}cos(4x)+\dfrac{4}{3}\dfrac{e^{3x}}{3}sen(4x)-\dfrac{16}{9}I+C\\\\\\I+\dfrac{16}{9}I=\dfrac{e^{3x}}{3}\left(cos(4x)+\dfrac{4}{3}sen(4x)\right)\\\\\\\dfrac{25}{9}I=\dfrac{e^{3x}}{3}\left(cos(4x)+\dfrac{4}{3}sen(4x)\right)\\\\\\\dfrac{25}{3}I=e^{3x}\left(cos(4x)+\dfrac{4}{3}sen(4x)\right)$


$\boxed{\boxed{I=\dfrac{3}{25}e^{3x}\left(cos(4x)+\dfrac{4}{3}sen(4x)\right)}}$