Question

Integral por partes : ∫e^x senx dx

Answer 1

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

Fazendo

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

Temos, pela fórmula da integração por partes:

$\displaystyle\int e^{x}sen(x)dx=uv-\int vdu\\\\\\\displaystyle\int e^{x}sen(x)dx=e^{x}sen(x)-\int e^{x}cos(x)dx$
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Agora, vamos achar uma expressão para a integral da direita, integrando por partes novamente:

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

Daí,

$\displaystyle\int e^{x}cos(x)dx=ab-\int bda\\\\\\\int e^{x}cos(x)dx=e^{x}cos(x)-\int e^{x}(-1)sen(x)dx\\\\\\\boxed{\boxed{\int e^{x}cos(x)dx=e^{x}cos(x)+\int e^{x}sen(x)dx}}$
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Substituindo essa expressão da integral na primeira:

$\displaystyle\int e^{x}sen(x)dx=e^{x}sen(x)-\int e^{x}cos(x)dx\\\\\\\int e^{x}sen(x)dx=e^{x}sen(x)-\left[e^{x}cos(x)+\int e^{x}sen(x)dx\right]\\\\\\\int e^{x}sen(x)dx=e^{x}sen(x)-e^{x}cos(x)-\int e^{x}sen(x)dx\\\\\\\int e^{x}sen(x)dx+\int e^{x}sen(x)dx=e^{x}\cdot[sen(x)-cos(x)]\\\\\\2\int e^{x}sen(x)dx=e^{x}\cdot[sen(x)-cos(x)]\\\\\\\boxed{\boxed{\int e^{x}sen(x)dx=\dfrac{e^{x}}{2}\cdot[sen(x)-cos(x)]+constante}}$