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

1 Answer
Lovecraft
Jan 7, 2016

#I = -x^2cos(x) + 2xsin(x) + 2cos(x) + c#

Explanation:

Say #u = x^2# so #du = 2x#, #dv = sin(x)# so #v = -cos(x)#

#I = -x^2cos(x) +2intxcos(x)dx#

Say #u = x# so #du = 1#, #dv = cos(x)# so #v = sin(x)#

#I = -x^2cos(x) + 2xsin(x) - 2intsin(x)dx#
#I = -x^2cos(x) + 2xsin(x) + 2cos(x) + c#

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