How would you evaluate this integral ?

x^2/(36-x^2)^(3/2)

Parts or Substitution ?(I tried using substitution) Please explain.

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How would you evaluate this integral ?

x^2/(36-x^2)^(3/2)

Parts or Substitution ?(I tried using substitution) Please explain.

x^2/(36-x^2)^(3/2)

Parts or Substitution ?(I tried using substitution) Please explain.

- chinex9a
**Posts:**10**Joined:**Mon Sep 13, 2010 4:42 am

I would suggest using substitution: x = 6*sin(u).

Wolfram: http://www.wolframalpha.com/input/?i=integrate+%28x^2%29%2F%2836-x^2%29^%283%2F2%29

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Wolfram: http://www.wolframalpha.com/input/?i=integrate+%28x^2%29%2F%2836-x^2%29^%283%2F2%29

(copy and paste the above into your browser's location bar)

- nona.m.nona
**Posts:**249**Joined:**Sun Dec 14, 2008 11:07 pm

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