# Tough Integral sqrt(x-sqrt(x^2-4))

• Mar 25th 2012, 11:16 AM
TheOnlyForgotten
Tough 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 PM
Prove It
Re: Tough Integral sqrt(x-sqrt(x^2-4))
Quote:

Originally Posted by TheOnlyForgotten
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!

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 PM
TheOnlyForgotten
Re: 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 PM
skeeter
Re: Tough Integral sqrt(x-sqrt(x^2-4))
Quote:

Originally Posted by TheOnlyForgotten
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.

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 PM
Prove It
Re: Tough Integral sqrt(x-sqrt(x^2-4))
Quote:

Originally Posted by skeeter
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

Wolfram DOES give me an answer...

integral&#91;Sqrt&#91;x - Sqrt&#91;x&#94;2 - 4&#93;&#93;&#93; - Wolfram|Alpha

But no steps...
• Mar 25th 2012, 06:06 PM
skeeter
Re: 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 AM
tom@ballooncalculus
Re: 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:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

... 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:
http://www.ballooncalculus.org/asy/prod.png

... is the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.

http://www.ballooncalculus.org/asy/maps/parts.png

... is lazy integration by parts, doing without u and v.

The rest...

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