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 $u=3x$ :

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

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