How do you find the antiderivative of #int sin^5(3x)cos(3x)dx#?

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Answer 1 · 2016-11-03T03:18:46

#sin^6(3x)/18+C#

#I=intsin^5(3x)cos(3x)dx#

First let #u=3x#. Thus #du=3dx# and we can modify the integral accordingly:

#I=1/3intsin^5(3x)cos(3x)(3dx)#

#I=1/3intsin^5(u)cos(u)du#

Now, let #v=sin(u)#. Differentiating shows that #dv=cos(u)du#. These are both already present:

#I=1/3intv^5dv#

Integrating using the typical power rule for integration:

#I=1/3 v^6/6+C#

Since #v=sin(u)#:

#I=sin^6(u)/18+C#

Since #u=3x#:

#I=sin^6(3x)/18+C#