Question

How do you integrate `x^3/sqrt(144-x^2)`?

Answer 1

Use substitution with `u = 144-x^2`.

Explanation:

`int x^3/sqrt(144-x^2) dx`

Let `u = 144-x^2`, so that `du = -2xdx` and `x^2 = 144-u`

`int x^3/sqrt(144-x^2) dx = int x^2/sqrt(144-x^2)" " xdx`

`= -1/2 int x^2/sqrt(144-x^2)" "(-2 x)dx`

`= -1/2 int (144- u)/sqrtu" "du`

`= -1/2 int (144/sqrtu - sqrtu)" "du`

`= -1/2 int (144u^(-1/2) -u^(1/2))" "du`

`= -1/2[288u^(1/2)-2/3u^(3/2)] +C`

`= -144(144-x^2)^(1/2)+ 1/3(144-x^2)^(3/2) +C`

`= -144sqrt(144-x^2)+ 1/3(sqrt(144-x^2))^3 +C`