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.

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

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

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