Question

PLEASE HELP ME ANSWER THIS QUESTION I NEED IT PLEASE.
Find ∫ x^5 √(2 - x^3) dx

Answer 1

`\displaystyle (2)/(15)(2-x^3)^{(5)/(2)}-(4)/(9)(2-x^3)^{(3)/(2)}+C`

Explanation:

`\displaystyle \int x^5√(2-x^3)\,dx\\\\=\int x^5(2-x^3)^{(1)/(2)}\,dx\\\\=\int x^3x^2(2-x^3)^{(1)/(2)}\,dx` <-- Breaking up
`x^5` into
`x^3x^2` helps us later

Let
`u=2-x^3` and
`du=-3x^2\,dx` so that
`2-u=x^3` and
`\displaystyle -(1)/(3)du=x^2dx`:

`\displaystyle -(1)/(3) \int (2-u)u^{(1)/(2)}\,du\\\\=-(1)/(3) \int 2u^{(1)/(2)}-u^{(3)/(2)}\,du\\\\=-(1)/(3)\biggr((4)/(3)u^{(3)/(2)}-(2)/(5)u^{(5)/(2)}\biggr)+C\\\\=-(1)/(3)\biggr((4)/(3)(2-x^3)^{(3)/(2)}-(2)/(5)(2-x^3)^{(5)/(2)}\biggr)+C\\\\=-(4)/(9)(2-x^3)^{(3)/(2)}+(2)/(15)(2-x^3)^{(5)/(2)}+C\\\\=(2)/(15)(2-x^3)^{(5)/(2)}-(4)/(9)(2-x^3)^{(3)/(2)}+C`