Integration by parts?

Question

$\int 6 x^{2} sin (3 x) d x$ $int 6x^2sin(3x)dx$

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Answer 1 · 2018-05-24T08:25:28

$- 2 x^{2} cos (3 x) + \frac{4 x sin (3 x)}{3} + \frac{4 cos (3 x)}{9} + C$ $-2x^2cos(3x)+(4xsin(3x))/3+(4cos(3x))/9+C$

First, let's factor out $6$ $6$ to leave us with $\int x^{2} sin (3 x) d x$ $intx^2sin(3x)dx$

$u'=sin(3x),u=-cos(3x)/3$
$v=x^2,v'=2x$

$6(-(x^2cos(3x))/3+2/3intxcos(3x)dx)$

$u'=cos(3x),u=sin(3x)/3$
$v=x,v'=1$

$6(-(x^2cos(3x))/3+2/3((xsin(3x))/3-intsin(3x)/3dx))$

$6(-(x^2cos(3x))/3+2/3((xsin(3x))/3+cos(3x)/9))$

$-2x^2cos(3x)+(4xsin(3x))/3+(4cos(3x))/9+C$