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How to integrate u^2/(sqrt(a^2-u^2)
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is that supposed to be an a^2??
Yes, it is very similar to 1\[\frac{ 1 }{ a ^{2}+x ^{2} } = \frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a }\] but is has the x^2 on top
It should be interesting this, \[\int\frac{u^2}{\sqrt{a^2-u^2}}du=\int\frac{a^2}{\sqrt{a^2-u^2}}du-\int\frac{a^2-u^2}{\sqrt{a^2-u^2}}du\] And I think it should be easy from this point.
How did you get you get that John_ES ?
are you doing trig substitution integrals yet?
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yes
then they probably want you to use the substitution\[u=\frac1a\sec\theta\]
for the record, John ES used the fact that u^2=a^2-(a^2-u^2), but I don't think that is the way to go with this problem
Thx, will try that this afternoon
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