How do you integrate #int x^2*sin(2x) dx# from #[0,pi/2]#?

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Answer 1 · 2018-04-12T16:13:22

The answer is #=pi^2/8-1/2#

First, calculate the indefinite integral by applying the integration by parts #2# times

#intuv'=uv-intu'v#

#u=x^2#, #=>#, #u'=2x#

#v'=sin2x#, #=>#, #v=-1/2cos2x#

The integral is

#I=intx^2sin2xdx=-x^2/2cos2x+intxcos2xdx#

#u=x#, #=>#, #u'=1#

#v'=cos2x#, #=>#, #v=1/2sin2x#

#intxcos2xdx=x/2sin2x-1/2intsin2xdx#

#=x/2sin2x+1/4cos2x#

Finally,

#I=-x^2/2cos2x+x/2sin2x+1/4cos2x#

And the definite integral is

#intx^2sin2xdx=[-x^2/2cos2x+x/2sin2x+1/4cos2x]_0^(pi/2)#

#=(pi^2/8-1/4)-(1/4)#

#=pi^2/8-1/2#

#=0.73#