How do you find the antiderivative of $int sin^3xdx$ ?

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Answer 1 · 2016-12-04T01:57:34

$int sin^3xdx = (cos^3x)/3-cosx+C$

Note that:

$sin^3x = sin x*sin^2x= sin x (1-cos^2x)$

So:

$int sin^3xdx = int sin x (1-cos^2x)dx = int sin x dx -int cos^2x sinxdx$

But $-sinx dx = d cosx$

$int sin^3xdx = -int d(cosx) +int cos^2x d( cosx ) = (cos^3x)/3-cosx$

How do you find the antiderivative of int sin^3xdxint sin^3xdx?