Question

Resolva a integra ∫ (x+1)cos2xdx pelo método da integração por partes

Answer 1

Fazendo as substituições:

$u=x+1~~~~~~~~~\longrightarrow~~~~~~~~~du=dx\\\\\frac{1}{2}sen(2x)~~~~~~~~~~\longleftarrow~~~~~~~~~v=cos(2x)dx$

Temos, pela integração por partes, que

$\displaystyle\int udv=uv-\int vdu\\\\\\\int(x+1)cos(2x)dx=(x+1)\dfrac{sen(2x)}{2}-\int\dfrac{1}{2}sen(2x)dx\\\\\\\int(x+1)cos(2x)dx=(x+1)\dfrac{sen(2x)}{2}-\dfrac{1}{2}\int sen(2x)dx$

Note que, se fizermos $a=2x$, então

$da=2dx~~\therefore~~dx=\frac{1}{2}da$

Daí,

$\displaystyle\int sen(2x)dx=\int sen(a)\dfrac{1}{2}da=\dfrac{1}{2}\int sen(a)da=-\dfrac{1}{2}cos(a)\\\\\\\boxed{\boxed{\int sen(2x)dx=-\dfrac{1}{2}cos(2x)}}~~(sem~a~constante)$
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Voltando:

$\displaystyle\int(x+1)cos(2x)dx=(x+1)\dfrac{sen(2x)}{2}-\dfrac{1}{2}\int sen(2x)dx\\\\\\\int(x+1)cos(2x)dx=(x+1)\dfrac{sen(2x)}{2}-\dfrac{1}{2}\left(-\dfrac{1}{2}cos(2x)\right)+C\\\\\\\boxed{\boxed{\int(x+1)cos(2x)dx=(x+1)\dfrac{sen(2x)}{2}+\dfrac{cos(2x)}{4}+C}}$