Question

Qual é o valor da integral indefinida $\int\limits {(x+2) cos3x} \, dx$ ?

Answer 1

Substituição de variáveis não resolveria a integral, então usamos integração por partes

Vamos fazer as seguintes sustituições:

$u=x+2\\dv=cos(3x)dx$

Então:

$u=x+2~~~~~~~~~~\therefore~~~~du=dx\\\\dv=cos(3x)dx~~~\therefore~~~v=\int cos(3x)dx=(\frac{1}{3})sen(3x)$

Integrando por partes:

$\displaystyle\int udv=uv-\int vdu\\\\\\\int(x+2)cos(3x)dx=(x+2)\dfrac{1}{3}sen(3x)-\int\dfrac{1}{3}sen(3x)dx\\\\\\\int(x+2)cos(3x)dx=\dfrac{x+2}{3}sen(3x)-\frac{1}{3}\int sen(3x)dx$

Integramos sen(3x) por substituição de variáveis

$a=3x~~~\therefore~~~da=3dx~~\rightarrow~~dx=(\frac{1}{3})da\\\\\\\displaystyle\int sen(3x)dx=\int sen(a)\frac{1}{3}da\\\\\\\int sen(3x)dx=\dfrac{1}{3}\int sen(a)da\\\\\\\int sen(3x)dx=-\dfrac{1}{3}cos(a)+C\\\\\\\boxed{\boxed{\int sen(3x)dx=-\dfrac{1}{3}cos(3x)+c_{1}}}$

Logo:

$\displaystyle\int(x+2)cos(3x)dx=\dfrac{x+2}{3}sen(3x)-\frac{1}{3}\left(-\dfrac{1}{3}cos(3x)+c_{1}\right)\\\\\\\boxed{\boxed{\int(x+2)cos(3x)dx=\dfrac{x+2}{3}sen(3x)+\dfrac{1}{9}cos(3x)+C}}$