Attachment 28533book answer is -.134

They use u=1-x^2

But what is wrong with what I did?

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- Jun 5th 2013, 12:33 PMminneola24Integral arcsin
Attachment 28533book answer is -.134

They use u=1-x^2

But what is wrong with what I did? - Jun 5th 2013, 12:54 PMPlatoRe: Integral arcsin
Is this the actual problem?

$\displaystyle \int_{\frac{-1}{2}}^0 {\frac{{xdx}}{{\sqrt {1 - {x^2}} }}} $ - Jun 5th 2013, 12:57 PMminneola24Re: Integral arcsin
- Jun 5th 2013, 01:04 PMHallsofIvyRe: Integral arcsin
By taking $\displaystyle v= x^2$, you wind up with $\displaystyle \int \frac{dv}{\sqrt{1- v}}$ NOT $\displaystyle \int\frac{dv}{\sqrt{1- v^2}}dv$

(Plato wasn't saying there was a difference, he was asking you to verify what your problem is.) - Jun 5th 2013, 01:35 PMminneola24Re: Integral arcsin
Yes he wrote it correctly. The book uses u=1-x^2

I'm just confused because I did it a different way and got a different answer. What did I do wrong? - Jun 5th 2013, 01:46 PMPlatoRe: Integral arcsin
- Jun 5th 2013, 01:49 PMminneola24Re: Integral arcsin
I understand you can do it without the u sub I did, but if let's say this is all that came to your mind at the time what did I do wrong in my work?

- Jun 10th 2013, 02:41 PMminneola24Re: Integral arcsin
Anyone have an idea?

Thanks - Jun 10th 2013, 06:02 PMtopsquarkRe: Integral arcsin