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

1 Answer

Oct 6, 2015

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

Explanation:

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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