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

Question

20002 views around the world

You can reuse this answer
Creative Commons License

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#