How do you integrate $int sin^2x$ by integration by parts method?

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Answer 1 · 2016-07-26T10:26:19

$I = 1/2(x - sin x cos x) + C$

so you have to use IBP on

$I = int sin^2x \ dx$

$= int sinx * sin x \ dx$

$= int sinx * d/dx (- cos x) \ dx$

which by IBP becomes...
$I =- sin x cos x + int d/dx (sinx) * cos x \ dx$

$=- sin x cos x + int cos^2 x \ dx$

$=- sin x cos x + int 1 - sin^2 x \ dx$

$=- sin x cos x + x - I + C$

$I = 1/2(x - sin x cos x) + C$

How do you integrate int sin^2xint sin^2x by integration by parts method?