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

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Answer 1 · 2016-04-21T14:58:03

#cos(x)-cos^3(x)/3+C#

We can integrate #sin^3(x)# if we use the Pythagorean identity to rewrite it:

#sin^3(x)=sin^2(x)sin(x)=(1-cos^2(x))sin(x)#

Thus, we see that

#intsin^3(x)dx=int(1-cos^2(x))sin(x)dx#

We can now integrate through substitution, since we have both something and its derivative present.

Set #u=cos(x)#, so that we also have #du=sin(x)dx#.

Substituting these both in, we see that

#intunderbrace((1-cos^2(x)))_(1-u^2)*underbrace(sin(x)dx)_(du)=int(1-u^2)du#

Integrating term by term, this gives us

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