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Try that and see what you get from there. Oh yeah, I had to a second substitution after the integration by parts in the second integral. | Try that and see what you get from there. Oh yeah, I had to a second substitution after the integration by parts in the second integral. | ||
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| + | *if you were going to do it the first way, I would think it should be ...<math>-\int_{0}^{\frac{1}{\sqrt{2}}}\frac{2x^3}{\sqrt{1-x^4}}dx</math>, because you forgot to take the du for sin^-1. | ||
== [[Page 563: Problem 17_MA181Fall2008bell]] == | == [[Page 563: Problem 17_MA181Fall2008bell]] == | ||
Revision as of 10:13, 14 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)
- I attempted to do the same thing at first with this problem as well. I couldn't get anywhere either. So I tried something totally different. I did a substitution before I integrated by parts.
let $ p=x^2,dp=2xdx $
Try that and see what you get from there. Oh yeah, I had to a second substitution after the integration by parts in the second integral.
- if you were going to do it the first way, I would think it should be ...$ -\int_{0}^{\frac{1}{\sqrt{2}}}\frac{2x^3}{\sqrt{1-x^4}}dx $, because you forgot to take the du for sin^-1.
