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 Posted: Thu Jan 01, 2009 7:27 am
 (both R and x under the root are squared)
I think it needs to use partial differentiation where Iudv = uv - Ivdu, but I can't figure out what to use for u and what for v.
Can anyone help?
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Posted: Mon Jan 19, 2009 11:00 pm
I have almost found the answer. The sqareroot came from finding the hight of a right triangle. I just have to find it using cosine and sine and I'll be ok, that'll be a solvable integral.
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Posted: Sat Mar 07, 2009 1:18 pm
Nop, that was not it. It required a trigonometric substitution.
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