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This looks better... but then I can't figure out how to solve that integral. Anyone? I've tried using the bottom as dv and going back to the inverse sin but that just comes up with | This looks better... but then I can't figure out how to solve that integral. Anyone? I've tried using the bottom as dv and going back to the inverse sin but that just comes up with | ||
| − | + | A = ##### + A and the A's cancel out, so that doesn't work.. Thanks. [[User:Idryg|Idryg]] 16:10, 13 October 2008 (UTC) | |
Revision as of 11:11, 13 October 2008
Page 552: Problem 20
- $ \int_{0}^{\frac{1}{\sqrt{2}}}2x\sin^{-1}(x^2)dx $
Using 2x as u gets rid of it eventually, but the inverse sin just gets worse and worse. Actually the integral of the inverse sin is just the inverse sin minus some radical. So it just cycles through over and over. So using the inverse sin looks better... It comes up with
$ [x^2\sin^{-1}(x^2)]_0^{\frac{1}{\sqrt{2}}} - \int_{0}^{\frac{1}{\sqrt{2}}}\frac{x^2}{\sqrt{1-x^4}}dx $
This looks better... but then I can't figure out how to solve that integral. Anyone? I've tried using the bottom as dv and going back to the inverse sin but that just comes up with A = ##### + A and the A's cancel out, so that doesn't work.. Thanks. Idryg 16:10, 13 October 2008 (UTC)
