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Integration cos^2(∏/2cosθ)

  1. Dec 3, 2012 #1
    i facing a maths problem in integrating ∫ cos^2(∏/2cosθ) with limit from 0 to ∏/2,i was panic and struggled a long period of time in solving this,anyone can help me? pls give me the answer in detail tq !
     
  2. jcsd
  3. Dec 3, 2012 #2
    Use the integral identity [itex]\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx[/itex].
     
  4. Dec 3, 2012 #3

    arildno

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    Why would that help, Millenial?
    You change the internal cos(theta) to a sin(theta)..
     
  5. Dec 3, 2012 #4

    micromass

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    First of all, do you mean

    (1) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta[/itex]

    or

    (2) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2\cos \theta}) d\theta[/itex]

    Also, which methods do you have at your disposal? Contour integration? Differentiation under the integral sign? Just normal calc II techniques?
     
  6. Dec 3, 2012 #5
    (1) [itex]\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta[/itex]
    ,i am asking this pls let me know the steps it takes
     
  7. Dec 3, 2012 #6

    micromass

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    Which course is this for? Do you know Bessel functions?? The solution requires this (at least that is what wolfram alpha says).
     
  8. Dec 3, 2012 #7

    arildno

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    Just a thought:
    We may easily rewrite this equation into the identity:
    [tex]\int_{0}^{\frac{\pi}{2}}\cos^{2}(\frac{\pi}{2}\cos\theta)d\theta+\int_{0}^{\frac{\pi}{2}}\sin^{2}( \frac{\pi}{2}\cos\theta)d\theta=\frac{\pi}{2}[/tex]
    I feel dreadfully tempted to declare the two integrals to have the same value (the latter being merely a flipped version of the first), but temptation is not proof..
     
    Last edited by a moderator: Dec 3, 2012
  9. Dec 3, 2012 #8

    micromass

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    Hmmm, looking at the graph doesn't really convince me that the integrals are equal :frown:

    Anyway, wolfram alpha gives us

    [itex]\int_0^{\pi/2} \cos^2(\frac{\pi}{2} \cos(x))dx = \frac{\pi}{4}(1+J_0(\pi))[/itex]

    so I doubt the integral will be solvable with methods like these.
     
    Last edited by a moderator: Dec 3, 2012
  10. Dec 3, 2012 #9

    Mute

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    To evaluate the integral one will have to use the identities

    $$\cos t = \frac{1}{2}(e^{it}+e^{-it})$$
    (or just ##\cos t = \mbox{Re}[\exp(it)]##)

    and

    $$e^{iz\cos\theta} = \sum_{n=-\infty}^\infty i^n J_n(z)e^{in\theta}.$$

    I guess the trig identity

    $$\cos^2 t = \frac{1}{2}(1+\cos(2t))$$
    also helps.
     
    Last edited: Dec 3, 2012
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