Ok so I tried U-subbing sqrt(x^2-4), but I didn't get anywhere. Then I though hmm, maybe trig-sub will do the trick, but I also got stuck there. I don't know what's left to do!

edit:

The integral is:

sqrt(x-sqrt(x^2-4))

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- Mar 25th 2012, 12:16 PMTheOnlyForgottenTough Integral sqrt(x-sqrt(x^2-4))
Ok so I tried U-subbing sqrt(x^2-4), but I didn't get anywhere. Then I though hmm, maybe trig-sub will do the trick, but I also got stuck there. I don't know what's left to do!

edit:

The integral is:

sqrt(x-sqrt(x^2-4)) - Mar 25th 2012, 06:01 PMProve ItRe: Tough Integral sqrt(x-sqrt(x^2-4))
- Mar 25th 2012, 06:16 PMTheOnlyForgottenRe: Tough Integral sqrt(x-sqrt(x^2-4))
Sorry, my mistake Prove it, but the whole equation is in the title which is sqrt(x-sqrt(x^2-4)).

sqrt(x^2-4) is only part of the integral.

Again, I need to find the integral of sqrt(x-sqrt(x^2-4)). Thanks. - Mar 25th 2012, 06:31 PMskeeterRe: Tough Integral sqrt(x-sqrt(x^2-4))
what makes you think this integral has a closed-form antiderivative in the first place?

Wolfram's result ...

integrate sqrt(x - sqrt(4-x^2)) dx - Wolfram|Alpha - Mar 25th 2012, 06:35 PMProve ItRe: Tough Integral sqrt(x-sqrt(x^2-4))
Wolfram DOES give me an answer...

integral[Sqrt[x - Sqrt[x^2 - 4]]] - Wolfram|Alpha

But no steps... - Mar 25th 2012, 07:06 PMskeeterRe: Tough Integral sqrt(x-sqrt(x^2-4))
I suffered mathematical dyslexia ...

put instead of - Mar 26th 2012, 05:26 AMtom@ballooncalculusRe: Tough Integral sqrt(x-sqrt(x^2-4))
A shame it can't remember, but what Wolfram probably did is notice that the derivative of the integrand is virtually that integrand divided by

...

http://www.ballooncalculus.org/draw/...wice/seven.png

... where (key in spoiler) ...

__Spoiler__:

... and therefore do integration by parts (twice) on the product, ...

http://www.ballooncalculus.org/draw/...ice/sevena.png

... (key in spoiler) ...

__Spoiler__:

The rest...

__Spoiler__:

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