Question

Question #76bf7

Answer 1

`2x^3sin^-1x - (2/3)sqrt(1-x^2)(x^2+2)`

Explanation:

integration by parts formula:
`intu(dv)/dx` = `uv-int(vu')dx`

So pick something you can integrate (v), and something you can differentiate (u)
`u= sin^-1 x :. u' = 1/sqrt(1 - x^2`
`v=2x^3` because `v' = 6x^2`

then plug into the formula:

`sin^-1x*2x^3 - int(2x^3)/sqrt(1-x^2)dx`

`(2x^3)/sqrt(1-x^2)` is a product - can also write as `2x^3*(1-x^2)^(-1/2)` so will need to use integration by parts again. This comes out as `-(2/3)sqrt(1-x^2)(x^2+2)`