How do you integrate #int x^2 cos3 x dx # using integration by parts?

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Answer 1 · 2016-01-09T01:49:51

#I = (9x^2sin(3x) +18xcos(3x) -2sin(3x))/9 + c#

#I = intx^2cos(3x)dx#

Say #u = x^2# so #du = 2x# and #dv = cos(3x)# so #v = sin(3x)/3#

#I = (x^2sin(3x))/3 - 2/3intxsin(3x)dx#

Say #u = x# so #du = 1# and #dv = sin(3x)# so #v = -cos(3x)/3#

#I = (x^2sin(3x))/3 - 2/3(-xcos(3x) +1/3intcos(3x)dx)#

#I = (x^2sin(3x))/3 - 2/3(-xcos(3x) +sin(3x)/9) + c#

#I = (9x^2sin(3x) +18xcos(3x) -2sin(3x))/9 + c#