Integration by parts?

Question

#int 6x^2sin(3x)dx#

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Answer 1 · 2018-05-24T08:25:28

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

First, let's factor out #6# to leave us with #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#