Question

Calcule a integral por partes

Answer 1

Olá lucas!

Pela regra do "LIATE"

u = x
dv = Cos3xdx
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Derivando "x implicitame em relação a u"

$\\ u = x \\ \\ \frac{d}{du} (u) = \frac{d}{du} (x) \\ \\ 1 = 1\frac{dx}{du} \\ \\ 1du = 1dx \\ \\ du = dx$
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Achando nosso "dv"

$dv = Cos(3x)dx$

Integrando ambos os lados, teremos:

$\\ \int\limits dv {} \, = \int\limits Cos(3x)dx{} \, \\ \\ v = \int\limits Cos(3x)dx{} \,$

Recorrendo a seguinte formula de integral:

$\\ \int\limits Cos(kx) {} \, dx = \frac{Sen(kx)}{k}$

Como nesse caso, k = 3:

Então,

$\\ \int\limits Cos(3x)dx{} \, = \frac{Sen(3x)}{3}$
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Substituindo na integral por partes:

$\\ \int\limits udv{} \,= uv-\int\limits vdu{} \, \\ \\ \int\limits udv{} \,= \frac{xSen(3x)}{3} - \int\limits { \frac{Sen(3x)}{3} } \, dx$
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Aplicando outra formula para a segunda integral:

$\\ \int\limits Sen(kx){} \,dx = - \frac{Cos(kx)}{k}$

Então,

$\\ \int\limits \frac{Sen(3x)}{3} {} \, dx = \frac{1}{3} \int\limits Sen(3x) {} \, dx \\ \\ \frac{1}{3} \int\limits Sen(3x) {} \, dx = \frac{1}{3} [ - \frac{Cos(3x)}{3} ] \\ \\ \frac{1}{3} \int\limits Sen(3x) {} \, dx = - \frac{Cos(3x)}{9}$
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Substituindo na nossa integral maior essas informações:

$\\ \int\limits {xCos(3x)} \, dx = \frac{xSen(3x)}{3} - \int\limits \frac{Sen(3x)}{3} \, dx = \\ \\ \int\limits {xCos(3x)} \, dx = \frac{xSen(3x)}{3} - [- \frac{Cos(3x)}{9} ] \\ \\ \int\limits {xCos(3x)} \, dx = \frac{xSen(3x)}{3} + \frac{Cos(3x)}{9} \\ \\ \int\limits {xCos(3x)} \, dx = \frac{1}{3} [ xSen(3x) + \frac{Cos(3x)}{3}]$

Colocando nossa constante "K"

$\\ \int\limits {xCos(3x)} \, dx = \frac{1}{3} [ xSen(3x) + \frac{Cos(3x)}{3}] +K$

Answer 2

$\sf \displaystyle \int \:x~cos \left(3x\right)dx\\\\\\=\int \frac{u~cos \left(u\right)}{9}du\\\\\\=\frac{1}{9}\cdot \int \:u~cos \left(u\right)du\\\\\\=\frac{1}{9}\left(u~sin \left(u\right)-\int sin \left(u\right)du\right)\\\\\\=\frac{1}{9}\left(u~sin \left(u\right)-\left(-cos \left(u\right)\right)\right)\\\\\\=\frac{1}{9}\left(3x~sin \left(3x\right)-\left(-cos \left(3x\right)\right)\right)\\\\\\=\frac{1}{9}\left(3x~sin \left(3x\right)+cos \left(3x\right)\right)$

$\to \boxed{\sf =\frac{1}{9}\left(3x~sin \left(3x\right)+cos \left(3x\right)\right)+C}$