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Trig sine substitution doesnt work

  1. Jan 12, 2006 #1
    how would on integrate [tex] \int_{0}^{1} \sqrt{1 + 4t^2} dt [/tex]
    trig sub sittution doesnt work since one doesnt get tan^2 +1 .

    i tryed solving this with matematica and it yielded something with a sinh argument. I am not familiarwi the hyp sine substitution.
     
  2. jcsd
  3. Jan 12, 2006 #2

    George Jones

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    Use the sustitution [itex]2t = tan \theta[/itex]. Be prepared to integrate [itex]sec^3 \theta[/itex].

    Regards,
    George
     
  4. Jan 12, 2006 #3

    CarlB

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    You may find that a substitution of t for sinh(theta)/2 will assist. And then you might find it useful to know how to write cosh(theta) in terms of exponentials.

    Lots of ways to skin a small furry critter.

    Carl
     
  5. Jan 12, 2006 #4

    GCT

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    to make things appear simpler factor out the 4, then make the proper trig substitution; if you still don't get it, you've definitely got some more studying to do.
     
  6. Jan 12, 2006 #5

    George Jones

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    AbsolutelY!

    Another way is to make the substitution [itex]2t = sinh x[/itex], and then to make a "half-angle" substitution for [itex]cosh^{2} x[/itex].

    Regards,
    George
     
  7. Jan 12, 2006 #6
    would the limits of integration change in this process?
    for t=0, theta = 0
    for t =1 , theta = arctan 2
    is that right?
     
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