How do you find the antiderivative of #(sinx)^3#?

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Answer 1 · 2016-11-09T11:09:38

#=(cos^3x)/3-cosx+C# #" "# C is a constant.

#int(sinx)^3dx#

#intsinx(sinx)^2dx#

Let #color(red)(u = cosx" " )#then #" "du=-sinxdx" "rArr color(red)(sinxdx = -du)#

Knowing the trigonometric identity:

#cos^2x + sin^2x =1#

#sin^2x=1 - cos^2x#

#int(-du)sin^2x#

#=int-(1-cos^2x)du#

#=int-(1-u^2)du#

#=intu^2-1du#

#=intu^2du-int1du#

#=u^3/3-u+C#

Substituting #color(red)(u=cosx)#

#=(cos^3x)/3-cosx+C# #" "# C is a constant.