What is the antiderivative of #x(sinx)^2#?

Question

2408 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-04T21:13:32

#1/4x^2-1/4xsin2x+1/8cos2x#

#I = intxsin^2xdx#

#cos2x= cos^2-sin^2x , rArr sin^2x= (1-cos2x)/2 #

#I = intxsin^2xdx = int(x(1-cos2x))/2dx#
#I = 1/2intxdx- 1/2intxcos2xdx#

Let evaluate the second part first. Let

#u = 2x rArr du = 2dx# and perform integration by parts

#1/2intxcos2xdx = 1/8 intucosudu= 1/8 (usinu-intsinudu)#
#=1/8 (usinu+cosu)#

#I = 1/4x^2-1/8(2xsin2x+cos2x)#
#I= 1/4x^2-1/4xsin2x+1/8cos2x#