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, 11:16 AMTheOnlyForgottenTough 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, 05:01 PMProve ItRe: Tough Integral sqrt(x-sqrt(x^2-4))
Hyperbolic substitution will work... You should know that $\displaystyle \displaystyle \begin{align*} \cosh^2{t} - \sinh^2{t} \equiv 1 \implies \cosh^2{t} - 1 &\equiv \sinh^2{t} \end{align*}$, so the substitution $\displaystyle \displaystyle \begin{align*} x = 2\cosh{t} \implies dx = 2\sinh{t}\,dt \end{align*}$ will work.

$\displaystyle \displaystyle \begin{align*} \int{\sqrt{x^2 - 4}\,dx} &= \int{\sqrt{\left(2\cosh{t}\right)^2 - 4}\cdot 2\sinh{t}\,dt} \\ &= \int{\sqrt{4\cosh^2{t} - 4}\cdot 2\sinh{t}\,dt} \\ &= \int{\sqrt{4\left(\cosh^2{t} - 1\right)} \cdot 2\sinh{t}\,dt} \\ &= \int{\sqrt{4\sinh^2{t}}\cdot 2\sinh{t}\,dt} \\ &= \int{2\sinh{t} \cdot 2\sinh{t} \,dt} \\ &= 4\int{\sin^2{t}\,dt} \\ &= 4\int{\frac{1}{2}\cosh{2t} - \frac{1}{2}\,dt} \\ &= 4\left(\frac{1}{4}\sinh{2t} - \frac{1}{2}t\right) + C \\ &= \sinh{2t} - 2t + C \\ &= 2\sinh{t}\cosh{t} - 2t + C \\ &= 2\cosh{t}\sqrt{\cosh^2{t} - 1} - 2t + C \\ &= x\sqrt{\left(\frac{x}{2}\right)^2 - 1} - \textrm{arcosh}\,{\frac{x}{2}} + C \\ &= x\sqrt{\frac{x^2 - 4}{4}} - \textrm{arcosh}\,{\frac{x}{2}} + C \\ &= \frac{x\sqrt{x^2 - 4}}{2} - \textrm{arcosh}\,{\frac{x}{2}} + C \end{align*}$ - Mar 25th 2012, 05:16 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, 05:31 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, 05:35 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, 06:06 PMskeeterRe: Tough Integral sqrt(x-sqrt(x^2-4))
I suffered mathematical dyslexia ...

put $\displaystyle \sqrt{4-x^2}$ instead of $\displaystyle \sqrt{x^2-4}$ - Mar 26th 2012, 04:26 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

$\displaystyle \sqrt{x^2 - 4}$...

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

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

__Spoiler__:

... and therefore do integration by parts (twice) on the product, $\displaystyle \frac{\sqrt{x - \sqrt{x^2 - 4}}}{\sqrt{x^2 - 4}}\ \sqrt{x^2 - 4}$ ...

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

... (key in spoiler) ...

__Spoiler__:

The rest...

__Spoiler__:

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