How do you find the integral of $sin^2x cosx$ ?

Question

101719 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-05T01:15:53

$int sin^2xcosxdx = 1/3sin^3x + C$

Let $I = int sin^2xcosxdx$

We can integrate this by substitution:

Let $u=sinx => (du)/dx=cosx$
$=> int ... cosxdx = int ... du$

And so, we can rewrite the integral as follows:
$I = int (sin^2x)cosxdx$
$I = int u^2du$
$I = 1/3u^3$

Ad re-substituting for $u$ we gte:
$I = 1/3sin^3x + C$

How do you find the integral of sin^2x cosx sin^2x cosx ?