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Inverse Trig Function Equivalence in Integral

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate ∫(x^2-4)^(1/2) / x for x > 2

    2. Relevant equations



    3. The attempt at a solution

    I was able to solve this problem via substitution, and my answer is: (x^2-4)^(1/2) - 2arcsec(x/2) + C. However, when I put the question into Wolfram Alpha, it gets this answer: (x^2-4)^(1/2) + 2arctan(2/(x^2-4)^(1/2)) + C. When I ask Wolfam Alpha to show the steps to the solution, it's second to last step is exactly the same as my answer, and then it says "which is equivalent for restricted x values to:" and shows it's final answer as I just typed. I am confused about how those two answers are the same. It seems to be saying that 2arctan(2/(x^2-4)^(1/2)) and -2arcsec(x/2) are equivalent (+/- a constant as C could have changed). How is this so?

    Link to Wolfram Alpha solution: http://www.wolframalpha.com/input/?i=integrate+(x^2-4)^(1/2)/xdx

    Thank you very much for your help! :)
     
  2. jcsd
  3. Sep 7, 2013 #2

    vela

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    Set up a right triangle with a hypotenuse of length x and the leg opposite the angle ##\theta## having length 2. What are ##\sec \theta## and ##\tan \theta## equal to?
     
  4. Sep 7, 2013 #3
    Thank you! I understand better now. One more thing though, why does the sign change? it was a negative inverse secant, but the arctan is positive.
     
  5. Sep 7, 2013 #4

    vela

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    Oops, I suggested the wrong triangle. You want the adjacent leg to be length 2, not the opposite leg. The minus sign comes from the fact that ##\arctan (x/y) = \pi/2 - \arctan(y/x)##.
     
  6. Sep 8, 2013 #5
    Thank you again! That makes perfect sense now. I realized now with your first triangle I also made a mistake and calculates arccsc instead of arcsec so it looked good.

    Why does Wolfram Alpha give the answer with arctan instead of arcsec? Is it preferred to do it that way? Also how do you keep your trig so sharp? I haven't done any math for a couple years and am now taking a differential equations class so I need to catch up on a lot of things. Thank you so much!
     
  7. Sep 8, 2013 #6

    vela

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    I don't know why Wolfram Alpha does the things it does. There's no reason to prefer arctan over arcsec, other than perhaps arctan might be a bit more familiar to most people.

    I've done this stuff for many many years, so I've had a lot of practice. Certain techniques turn out to be useful, and you remember them. A lot of the stuff you learned before and have pretty much forgotten you'll end up learning again. It'll be easier the second time around, and it'll tend to stick because you'll see how everything fits together.
     
  8. Sep 9, 2013 #7

    vanhees71

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    Hm, I'd use the standard substitution
    [tex]x=4 \cosh u[/tex]
    here ;-).
     
  9. Sep 9, 2013 #8
    Vanhees, could you explain how that would work?
     
  10. Sep 10, 2013 #9

    vanhees71

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    Ok. We want to find
    [tex]I=\int \mathrm{d} x \frac{\sqrt{x^2-4}}{x}.[/tex]
    Now I substitute
    [tex]x=2 \cosh u, \quad \mathrm{d}x=\mathrm{d} u 2 \sinh u,[/tex]
    leading to
    [tex]I=2\int \mathrm{d} u \frac{\sinh^2 u}{\cosh u}=2 \int \mathrm{d} u \left (\cosh u + \frac{1}{\cosh u} \right),[/tex]
    which is not so hard to integrate. Note that for the second you can use
    [tex]\cosh u=\cosh^2(u/2)+\sinh^2(u/2)=\cosh^2(u/2) [1+\tanh^2(u/2)].[/tex]
     
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