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Homework Help: Trigonometric integral

  1. Mar 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Integrate at interval [0,T] (T and k are given real numbers) the

    2. Relevant equation

    [tex]_{0}^{T}\int \frac{sin(p)}{\sqrt{k+p}}\ dp[/tex]

    3. The attempt at a solution

    [tex]\ Using\ substitution\ u\ =\ tan(p/2),\ results\ as\ :\ p\ =\ 2*arctan(u)\ \ ;\ \ dp\ =\ \frac{2}{1+u^2}\ du\ ;\ [/tex]

    [tex]sin(p)\ =\ \frac{2*u}{1+u^2} ;\ cos(p)\ =\ \frac{1-u^2}{1+u^2} ;\ [/tex]

    [tex]_{0}^{T}\int \frac{sin(p)}{\sqrt{k+p}}\ dp \ \ =\ _{0}^{2*arctan(T)}\int \frac{2*u*2}{(1+u^2)\ *\ \sqrt{k+2*arctan(u)}\ *\ (1+u^2)}\ du[/tex]

    [tex]\ ¿\ Could\ someone\ get\ a\ better\ result\ ?[/tex]

    (maybe with the substitution u = 2* sin(p) )


    ...
     
  2. jcsd
  3. Mar 19, 2010 #2
    See e.g. 16 http://mathworld.wolfram.com/FresnelIntegrals.html" [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Mar 20, 2010 #3
    Should be:

    [tex]
    _{0}^{T}\int \frac{sin(p)}{\sqrt{k+p}}\ dp \ \ =\ _{0}^{tan(T/2)}\int \frac{2*u*2}{(1+u^2)\ *\ \sqrt{k+2*arctan(u)}\ *\ (1+u^2)}\ du
    [/tex]

    but it looks like I could have to know about Fresnel Integrals. Thank you for the help.


    .
     
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