Integrate [x/(a^3 - x^3)]^1/2 dx.
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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.`
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