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

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Answer 1 · 2018-03-20T15:57:19

The answer is #=1/6x^3-x^2/4sin2x-x/4cos2x-1/8sin2x+C#

Reminder

#cos2x=1-2sin^2x#

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

#intuv'=uv-intu'v#

The integral is

#I=intx^2sin^2xdx=1/2intx^2(1-cos2x)dx#

#=1/2intx^2-1/2intx^2cos2xdx#

#=1/6x^3-1/2intx^2cos2xdx#

Perform the second integral by parts

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

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

So,

#intx^2cos2xdx=x^2/2sin2x-intxsin2xdx#

Calculate the third intehral by parts

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

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

So,

#intxsin2xdx=-x/2cos2x+1/2intcos2xdx#

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

Putting it all together

#I=1/6x^3-x^2/4sin2x-x/4cos2x-1/8sin2x+C#