# Integrate [x/(a^3 - x^3)]^1/2  dx.

sciencesolve | Certified Educator

calendarEducator since 2011

starTop subjects are Math, Science, and Business

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 yieldsint (sqrt x)/sqrt(a^3-x^3)dx = (2/(3a^(3/2))) arcsin((sqrt(x^3))/sqrt(a^3)) + c.

check Approved by eNotes Editorial