Question

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

Answer 1

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

Explanation:

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`