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Thread: Tough Integral sqrt(x-sqrt(x^2-4))

  1. March 25th 2012, 11:16 AM #1
    TheOnlyForgotten
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    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))
    Last edited by TheOnlyForgotten; March 25th 2012 at 05:17 PM.
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  2. March 25th 2012, 05:01 PM #2
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    Re: Tough Integral sqrt(x-sqrt(x^2-4))

    Quote Originally Posted by TheOnlyForgotten View Post
    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 \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 \begin{align*} x = 2\cosh{t} \implies dx = 2\sinh{t}\,dt \end{align*} will work.

    \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*}
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  3. March 25th 2012, 05:16 PM #3
    TheOnlyForgotten
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    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.
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  4. March 25th 2012, 05:31 PM #4
    skeeter
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    Re: Tough Integral sqrt(x-sqrt(x^2-4))

    Quote Originally Posted by TheOnlyForgotten View Post
    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
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  5. March 25th 2012, 05:35 PM #5
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    Re: Tough Integral sqrt(x-sqrt(x^2-4))

    Quote Originally Posted by skeeter View Post
    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[Sqrt[x - Sqrt[x^2 - 4]]] - Wolfram|Alpha

    But no steps...
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  6. March 25th 2012, 06:06 PM #6
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    Re: Tough Integral sqrt(x-sqrt(x^2-4))

    I suffered mathematical dyslexia ...

    put \sqrt{4-x^2} instead of \sqrt{x^2-4}
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  7. March 26th 2012, 04:26 AM #7
    tom@ballooncalculus
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    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

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



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

    Spoiler:


    ... 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, \frac{\sqrt{x - \sqrt{x^2 - 4}}}{\sqrt{x^2 - 4}}\ \sqrt{x^2 - 4} ...





    ... (key in spoiler) ...

    Spoiler:


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



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


    The rest...

    Spoiler:




    Now solve the top row for I.

    Full size


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    Last edited by tom@ballooncalculus; March 26th 2012 at 12:20 PM.
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