You should use the following substitution such that:

`sqrt(x^3) = t => (3x^2)/(2sqrt(x^3))dx = dt => (3x^2)/(2xsqrtx) dx = dt => (3x)/(2sqrtx) = dt `

Changing the variable yields:

`int ((sqrt x)/(sqrt(a^3-x^3)))dx = int (2dt)/(3sqrt(a^3 - t^2))`

You need to factor out `a^3` to denominator such that:

`(2/3)int (dt)/(a^(3/2)sqrt(1 - t^2/a^3))`

You should substitute u for `t/sqrt(a^3)` such that:

`u= t/sqrt(a^3) => du = a^(-3/2)dt`

Changing the variable yields:

`(2/3)int (dt)/(a^(3/2)sqrt(1 - t^2/a^3)) = ((2/3a^(3/2)))int (du)/(sqrt(1 - u^2))`

`(2/(3a^(3/2)))int (du)/(sqrt(1 - u^2)) = (2/(3a^(3/2))) arcsin u + c`

Substituting back `t/sqrt(a^3)` for u yields:

`(2/3)int (dt)/(a^(3/2)sqrt(1 - t^2/a^3)) = (2/(3a^(3/2))) arcsin(t/sqrt(a^3)) + c`

`int (sqrt x)/sqrt(a^3-x^3)dx = (2/(3a^(3/2))) arcsin((sqrt(x^3))/sqrt(a^3)) + c`

**Hence, evaluating the given integral yields`int (sqrt x)/sqrt(a^3-x^3)dx = (2/(3a^(3/2))) arcsin((sqrt(x^3))/sqrt(a^3)) + c.` **