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Integration using arcsin

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-05-27at101440PM.png



    3. The attempt at a solution

    I'm having a tough time understanding this step. I understand that

    the integral of 1/sqrt(a^2-x^2) = arcsin(u/a) + C

    But sqrt(4-x^2) does not have a 1 in the numerator. It looks like they're using u substitution, if so, then why is it + 4 arcsin rather than times 4arcsin. I also don't understand where the 4 comes from.
     
  2. jcsd
  3. May 27, 2012 #2
    I'm guessing it's just a typical trig sub. Set ##x = 2 \sin \theta##, use a trigonometric identity, integrate and then solve for ##\theta## in terms of x.

    If you haven't seen this technique before, check out this link.
     
  4. May 27, 2012 #3

    SammyS

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    What's the derivative of
    [itex]\displaystyle \frac{3}{2}\left(x\sqrt{4-x^2\,} -4\,\sin^{-1}\left(\frac{x}{2} \right) \right)\ ?[/itex]​
     
    Last edited: May 28, 2012
  5. May 27, 2012 #4
    If I knew I wouldn't have posted the thread
     
  6. May 27, 2012 #5
    well, if If the integrand contains x^2 − a^2, let

    [tex] x = a \sec \theta\,[/tex]

    and use the identity

    [tex] \sec^2 \theta-1 = \tan^2 \theta.\,[/tex]

    Then why do they use arcsin instead of secant
     
  7. May 27, 2012 #6

    SammyS

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    How do you expect to be able to integrate, if you don't know basic differentiation ?

    According to your OP, the derivative of [itex]\displaystyle \sin^{-1}\left(\frac{x}{2} \right) [/itex] is [itex]\displaystyle \frac{1}{\sqrt{4-x^2}}\ .[/itex]

    The rest is pretty basic.
     
  8. May 27, 2012 #7
    There's a very important difference in whether you have x2 - a2 or a2 - x2 so that you use x = asecθ or x = asinθ, respectively.
     
  9. May 28, 2012 #8
    well, is a^2-x^2 which results in asin. However, that integral only works if there is a one in the numerator and clearly there is not, so I'm still wondering where 4asin came from.

    Sammy, the rest is not basic.
     
  10. May 28, 2012 #9

    SammyS

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    How is [itex]\displaystyle \frac{d}{dx}\left(x\sqrt{4-x^2\,}\right)[/itex] not basic ?
     
  11. May 28, 2012 #10
    It's basic alright, but it doesn't even appear on the left side of the equation. Why is the text using that derivative?
     
  12. May 28, 2012 #11
    ∫√(4 - x2) dx

    x = 2sinθ → x2 = 22sin2θ
    dx = 2cosθ dθ

    [tex]\int\sqrt{4-4\sin^2\theta}\text{ }2\cos\theta\text{ }d\theta = \int\sqrt{4(1-sin^2\theta)}\text{ }2\cos\theta\text{ }d\theta = \int\sqrt{4\cos^2\theta}\text{ }2\cos\theta\text{ }d\theta[/tex][tex]= \int 2\cos\theta\text{ }2\cos\theta\text{ }d\theta = 4\int \cos^2\theta\text{ }d\theta = 4\int\frac{1}{2}(\cos2\theta + 1)\text{ }d\theta[/tex]The cos2θ part will turn into x√(4 - x2) and the 1 part will turn into sin-1(x/2) after substituting back for x and simplifying.
     
  13. May 28, 2012 #12
    robert, the method I outlined in my original response (which I've quoted above) is still correct. You had the right idea, but as Bohrok pointed out, the substitution should be ##x = 2 \sin \theta## (as I said originally). You can then use the identity ##1 - \sin^2 \theta = \cos^2 \theta##. What does the integral become with this substitution?

    Edit: Bohrok seems to have done a lot of the work for you. :grumpy:

    Note that he or she did leave out the factor of 3, however.
     
  14. May 28, 2012 #13

    SammyS

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    The reason I asked this question is that this derivative should be equal to [itex]\displaystyle 3\sqrt{4-x^2}\ .[/itex]

    The steps involved in taking this derivative might help you understand how one finds the anti-derivative.
     
  15. May 28, 2012 #14
    How is this legal?

    By what rule? I don't see how you can get cos to turn into x
     
  16. May 28, 2012 #15
    The book says they're equal so I already know that.

    That's what I'm trying to figure out.
     
  17. May 28, 2012 #16
    Legal? It's a substitution. If I said "Let ##\theta = \arcsin\left(x/2\right)##," would you still take issue with it? Certainly there is a lot of work you'd have to do to provide a rigorous proof that it works, but I'm guessing you've done substitutions before without taking a real analysis course. Take a look at the link I posted above.

    The last step that Bohrok did uses a double-angle formula. See if you can work it out. Once you've integrated (in terms of ##\theta##) substitute back in for x.
     
    Last edited: May 28, 2012
  18. May 28, 2012 #17
    We're driven to use x = 2sinθ since the domain of √(4 - x2) is [-2, 2], and the range of 2sinθ is [-2, 2]. Not tired just yet, so I typed up the steps (ignoring +C)

    [tex]x = 2\sin\theta \Longrightarrow \frac{x}{2} = \sin\theta \Longrightarrow\sin^{-1}\left(\frac{x}{2}\right) = \theta[/tex][tex]\int\cos2\theta\text{ }d\theta = \frac{1}{2}\sin2\theta = \frac{1}{2}\cdot2\sin\theta\cos\theta = \sin\theta\cos\theta = \frac{x}{2}\cos\left(\sin^{-1}\frac{x}{2}\right) = \frac{x}{2}\sqrt{1 - \left(\frac{x}{2}\right)^2} = \frac{x}{2}\sqrt{1 - \frac{x^2}{4}}[/tex]
    There's a trig identity up there where cos(sin-1x) = √(1 - x2)
     
  19. May 28, 2012 #18
    I see how they get cos^2 = cos 2a + 1

    But what I do not understand is how they get

    x = 2sin(a)
     
  20. May 28, 2012 #19
    If this substitution is causing much confusion you could try to integrate by parts. It may at least clear up why there are two terms.
     
  21. May 28, 2012 #20
    Bohrok,

    Thanks for your detailed help. I have never encountered these substitution techniques? Should I have encountered them by now? I'm up to triple integrals in calc. Is one expected to know these techniques by the time they reach triple integrals?
     
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