How do you find the integral of #x sin^2(x)#?

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Answer 1 · 2017-02-12T12:08:16

#intxsin^2xdx=x/4(2x-sin2x)-1/4x^2-1/8cos2x+C#

In order to integrate this function, we must integrate by parts, or use the reverse product rule, as I like to call it.

I won't go into much detail here about how to do that, but watch Khan academy's video for more help.

#intxsin^2xdx#

#u=x rarr(du)/dx=1#

#(dv)/dx =sin^2x=1/2(1-cos2x) rarr v =1/4(2x-sin2x) #

#intxsin^2xdx=1/4x(2x-sin2x)-1/4int2x-sin2xdx=#

#x/4(2x-sin2x)-1/4x^2-1/8cos2x+C#