How do you integrate #int sin^3x# by integration by parts method?

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Answer 1 · 2016-08-02T15:39:55

#=1/3 cos^3 - cos x + C#

#int \ sin^3 x \ dx#

#= int \d/dx (-cos x) sin^2 x \ dx#

which by IBP

#=-cos x sin^2x + int \cos x d/dx( sin^2 x) \ dx#

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

#=-cos x sin^2 x + 2 int \d/dx ( - 1/3 cos^3 x) \ dx#

#=-cos x sin^2 x - 2/3 cos^3 x + C#

#=-cos x (1 - cos^2 x) - 2/3 cos^3 x + C#

#=-cos x + cos^3 x - 2/3 cos^3 x + C#

#=1/3 cos^3 - cos x + C#