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

Answer 1

Aurora Alvarado

#=(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.

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Answer 2

Christopher Agresta

To find the antiderivative of ( (\sin x)^3 ), you can use the trigonometric identity ( \sin^2 x = 1 - \cos^2 x ) to express ( (\sin x)^3 ) in terms of ( \cos x ). Then, apply integration by substitution or integration by parts to find the antiderivative. The result will be ( -\frac{1}{3} \cos^3 x + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.