What is the integral of #sin^5(x)#?

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Answer 1 · 2018-03-30T01:14:40

#intsin^5xdx=sin^2x-sinx-1/3sin^3x+C#

We want #intsin^5xdx#

Rewrite the integrand as #sinx(sin^2x)^2#:

#intsinx(sin^2x)^2dx#

Recall the identity #sin^2x+cos^2x=1#. The identity also tells us that

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

#intsinx(1-cos^2x)^2dx#

This can be solved using a simple substitution. Let

#u=cosx, du=-sinxdx, -du=sinxdx#

Rewrite the integral:

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

Integrate:

#-int(1-2u+u^2)=-(u-u^2+1/3u^3)+C=u^2-u-1/3u^3+C#

Rewriting in terms of #x# yields

#intsin^5xdx=sin^2x-sinx-1/3sin^3x+C#