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Integral question with i.

  1. Apr 30, 2013 #1
    This is not homework I was wondering if I had to integrate
    [itex] \frac{dx}{\sqrt{x^2-1}} [/itex]
    Instead of doing the normal trig substitution what if I used
    [itex] sin(u)^2-1=-cos(u)^2 [/itex] x=sin(u) dx=cos(u)du
    But when I make the substitution I will get a negative sign under the radical, could I just pull it out as i and then some how extract the real part. How would I do it with the i in their.
  2. jcsd
  3. Apr 30, 2013 #2

    Simon Bridge

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    You could but you have to keep it - unless you have any reason to believe that the solution is to do with the real part (why not the imaginary part?)

    OR, you could use ##x=\sec\theta## and ##1+\tan^2\theta = \sec^2\theta##
  4. Apr 30, 2013 #3
    If you want a faster way than a normal trig substituion, try a hyperbolic substitution. [itex] x=\cosh t [/itex]and [itex] dx=\sinh t dt [/itex] and use the identity [itex] \cosh^2{t}-\sinh^2{t}=1 [/itex].
  5. Apr 30, 2013 #4
    I would like to learn how to do this with complex numbers, ok if I leave the i in their, how would I transform it in the end to get my answer. isnt [itex] cos(ix)=cosh(x) [/itex] or maybe we could do something with eulers formula.
  6. May 1, 2013 #5

    Simon Bridge

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    No idea since I'd never do it that way - if x=sin(u) then u=arcsin(x) ... so you'll have an arcsin and an i in the final form to simplify out. But I don't think there are nice results from that integration anyway.
  7. May 1, 2013 #6


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    You have to be more careful with your approach to this integral. You can't just set ##x = \sin t## without thinking about it. There are several cases you have to consider.

    First, is x greater or less than 1?

    If it's less than 1, ##\sqrt{x^2-1}## is imaginary, so you can just pull a minus sign out and integrate ##-i/\sqrt{1-x^2}## using your trigonometric substitution, though you still have to be careful, because ##\sqrt{1-\sin^2t} = |\cos t|##, not just ##\cos t##, so you need to worry about the sign of t as well.

    Now, if x is greater than one, then in the change of variables ##x = \sin t ## , t is necessarily complex (or at least imaginary). Otherwise, ##\sin t \leq 1##. Your integral then has a square root of a potentially complex number, and you can't just start pulling out i's or minus signs. You may need to worry about branch cuts and so on, as well as the appropriate interpretation of the integral now that you are integrating over a non-real variable t. The hyperbolic substitutions avoid all of these issues, of course.
  8. May 1, 2013 #7

    Simon Bridge

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    The trig substitutions are taught as a kind of quick sleight-of-hand aren't they?
    The standard lessons don't mention this stuff - and don't normally need to because it tends to come out in the wash for the problems chosen.
    It tends to be when students try to extend the techniques beyond the lessons that it gets noticeable.
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