How do you find the antiderivative of $(cos(3x))^3$ ?

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Answer 1 · 2017-10-22T15:35:53

$1/3sin3x-1/9sin^3 3x+c$

$intcos^3 3xdx$

$=intcos3xcos^2 3xdx$

$=intcos3x(1-sin^2 3x)dx$

$I=int(cos3x-cos3xsin^2 3x)dx$

now
$d/(dx)(sin^3 3x)=3 xx3xxsin^2 3xcos3x=9sin^2 3xcos3x$

$I=1/3sin3x-1/9sin^3 3x+c$

How do you find the antiderivative of (cos(3x))^3(cos(3x))^3?