Question

integral de x. cos3dx

Answer 1

$\displaystyle\int x\cdot cos(3x)dx$

Integramos essa função por partes, fazendo

$u=x~~~\therefore~~~\boxed{\boxed{du=dx}}\\\\dv=cos(3x)dx$

Achando v:

$v=\displaystyle\int cos(3x)dx$

Sendo a = 3x, da = 3dx, então dx = (1/3)da:

$v=\displaystyle\int cos(a)\dfrac{1}{3}da\\\\\\v=\dfrac{1}{3}\int cos(a)da\\\\\\v=\dfrac{1}{3}sen(a)\\\\\\\boxed{\boxed{v=\dfrac{1}{3}sen(3x)}}$

(Ignorando a constante de integração, pois essa entrará na integração por partes)
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Portanto, temos:

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

Integrando por partes:

$\displaystyle\int udv=uv-\int vdu\\\\\\\int xcos(3x)dx=x\dfrac{1}{3}sen(3x)-\int\dfrac{1}{3}sen(3x)dx\\\\\\\int xcos(3x)dx=\dfrac{1}{3}xsen(3x)-\dfrac{1}{3}\int sen(3x)dx$

Como 3x = a, da = 3dx, logo dx = (1 / 3)da:

$\displaystyle\int xcos(3x)dx=\dfrac{1}{3}xsen(3x)-\dfrac{1}{3}\int\dfrac{1}{3}sen(a)da\\\\\\\int xcos(3x)dx=\dfrac{1}{3}xsen(3x)-\dfrac{1}{3}\cdot\dfrac{1}{3}\int sen(a)da\\\\\\\int xcos(3x)dx=\dfrac{1}{3}xsen(3x)-\dfrac{1}{9}(-cos(a))+C\\\\\\\boxed{\boxed{\int xcos(3x)dx=\dfrac{1}{3}xsen(3x)+\dfrac{1}{9}cos(3x)+C}}$