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How do you find the integral of #int sin^3(x)dx#?

Answer 1

Samantha Adkison

#int sin^{3}(x)\ dx=1/3 cos^{3}(x)-cos(x)+C#

Use the identity #sin^{2}(x)=1-cos^{2}(x)# and then let #u=cos(x)# so that #du = -sin(x)\ dx# and

#int sin^{3}(x)\ dx=int(1-cos^{2}(x))sin(x)\ dx=int (u^2-1)\ du#

#=u^3/3-u+C=1/3 cos^{3}(x)-cos(x)+C#

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Answer 2

Aurora Alvarado

To find the integral of ( \int \sin^3(x) , dx ), you can use the trigonometric identity ( \sin^3(x) = (\sin^2(x))\sin(x) ). Then, perform a substitution using ( u = \sin(x) ) and ( du = \cos(x) , dx ). This transforms the integral into ( \int u^2 , du ), which can be easily integrated.

Therefore, the integral of ( \int \sin^3(x) , dx ) equals ( -\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.