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

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Answer 1 · 2015-11-03T15:30:41

#intsin^2(x)dx=x/2-sin(2x)/4+c#,
where #c# is the constant of integration.

#intsin^2(x)dx=intfrac{d}{dx}(x)sin^2(x)dx#

#=xsin^2(x)-intxfrac{d}{dx}(sin^2(x))dx#

#=xsin^2(x)-intx(2sin(x)cos(x))dx#

#=xsin^2(x)-intxsin(2x)dx#

#=xsin^2(x)+1/2intxfrac{d}{dx}(cos(2x))dx#

#=xsin^2(x)+1/2[xcos(2x)-intfrac{d}{dx}(x)cos(2x)dx]#

#=xsin^2(x)+1/2xcos(2x)-1/2intcos(2x)dx#

#=xsin^2(x)+1/2xcos(2x)-sin(2x)/4+c#,
where #c# is the constant of integration.

#=x/2(cos(2x)+2sin^2(x))-sin(2x)/4+c#

#=x/2-sin(2x)/4+c#