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Integration substitution

  1. Feb 28, 2006 #1
    Am i being really dumb when struggling to do this?

    [tex] L = 4\sqrt{2}c \left \int_{0}^{\frac{\pi}{2}} \left \frac{dt}{\sqrt{1 + sin^2 t}} [/tex]

    Using substitution or otherwise show that

    [tex] L = 4c \left \int_{0}^{1} \left \frac{du}{\sqrt{1 - u^4}} [/tex]

    Its a small part of a question but its stopping me doing the rest. Anyone help me out? The limits im fine with, for the rest i get:

    [tex] u = \sin t \left \frac{du}{dt} = \cos t \left dt = \frac{du}{\cos t} \left so \left L = \frac{du}{\sqrt{1 - u^2}\cos t} \left ??? [/tex]
  2. jcsd
  3. Mar 1, 2006 #2


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    I think it should be a [itex]\sqrt {1 + u^2 } [/itex] in the denominator, in what you have so far. Then remember that everything has to stand in the new variable u, you can't have any t's anymore.
    But since [itex] u = \sin t[/itex], we have that [itex]\cos t = \sqrt {1 - \sin ^2 t} = \sqrt {1 - u^2 } [/itex].

    Can you finish it?
  4. Mar 1, 2006 #3


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    [tex] \int \frac{dt}{\sqrt{1+\sin^{2}t}} =F\left(x|-1\right) [/tex]

    ,where F(x|m) is http://documents.wolfram.com/mathematica/functions/EllipticF/" [Broken].

    Last edited by a moderator: May 2, 2017
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