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Integration: inverse trigonometric functions

  1. Sep 13, 2015 #1
    1. The problem statement, all variables and given/known data
    ∫(t/√(1-t4))dt

    2. Relevant equations

    ∫(du/√(a2 - u2)) = arcsin (u/a) + C

    ∫(du/(a2 + u2) = (1/a) arctan (u/a)

    ∫(du/(u√(u2 - a2))) = (1/a) arcsec (|u|/a)

    3. The attempt at a solution

    Screenshot_2015-09-13-14-47-36.png

    Edit: I meant to write u where t2 is
     
    Last edited: Sep 13, 2015
  2. jcsd
  3. Sep 13, 2015 #2

    Titan97

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    That is not a proper way to write an integral. You should express t in terms of u.
    ##dt=\frac{dt}{2t}=\frac{dt}{2\sqrt{u}}##.
    But this does not make the integral easy.
    Instead, have you tried expressing it as $$\frac{1}{2}\int\frac{1+t^2+1-t^2}{\sqrt{(1+t^2)(1-t^2)}}dt$$
     
  4. Sep 13, 2015 #3
    I edited under that I meant to replace t2 with u :) No, but I wouldn't know what to do with it anyway?
     
  5. Sep 13, 2015 #4

    stevendaryl

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    My guess is that that integral does not have a solution in terms of the usual functions.
     
  6. Sep 13, 2015 #5
    In the back of my book it says the answer is (1/2) arcsin (t2) + C, but I don't know how to get there :-/
     
  7. Sep 13, 2015 #6
    Screenshot_2015-09-13-15-19-06.png
    What can I do next?
     
  8. Sep 13, 2015 #7

    stevendaryl

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    Hmm. Is it possible that there is a typo in your original integral? I get that answer if the original integral was as shown on the left-hand side below:

    [itex]\int \dfrac{t dt}{\sqrt{1-t^4}} = \frac{1}{2} arcsin(t^2)[/itex]
     
  9. Sep 13, 2015 #8
    Yes that is the integral, did I not write it correctly in my original post? How do you get your equations to look like that?o0)
    Edit: I just saw I wrote 1 instead of t, so sorry!
     
  10. Sep 13, 2015 #9

    Titan97

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    Use LaTeX
     
  11. Sep 13, 2015 #10

    stevendaryl

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    To prove that, you make the substitution:

    [itex]t = \sqrt{u}[/itex]

    then the integral in terms of [itex]u[/itex] should be one of the ones you listed in your first post.
     
  12. Sep 13, 2015 #11
    Think I got it, thanks :smile:

    Screenshot_2015-09-13-15-36-37.png
     
  13. Sep 13, 2015 #12

    Mark44

    Staff: Mentor

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