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Homework Help: Trigonometric Integration / relationship between f(cosx) and f(sinx)

  1. Nov 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that:

    ∫f(sin(x))dx = ∫f(cos(x))dx

    where each integral is over the limits [0, [itex]\pi[/itex]/2], for a 'well behaved' function f.

    2. The attempt at a solution

    I have tried relating sin(x) and cos(x) and somehow rearrange one of the integrals to look like the other. Since:

    sin(x) = cos(x-[itex]\pi[/itex]/2)

    I can substitute this into the LHS integral and I am sure this is the correct way to start. It is at this point where I cannot think of any way of moving forward.

    Any help would be appreciated.
    Last edited: Nov 22, 2012
  2. jcsd
  3. Nov 22, 2012 #2
    You are heading the right way. I'll give you two tips. The first is, how are [itex]\displaystyle \int_{0}^{t} f(x)\,dx[/itex] and [itex]\displaystyle \int_{0}^{t} f(t-x)\,dx[/itex] related?

    The second tip is, think about the fundamental properties of sine and cosine. There is one particular property that will help you.
  4. Nov 22, 2012 #3
    Thank you for you reply Millennial.

    Using the property that cos(-x) = cos(x) I have got:

    int[f(sinx)]dx = int[f(cos(x-[itex]\pi[/itex]/2))] = int[f(cos([itex]\pi[/itex]/2-x))]

    Looking at your first tip one would assume that the relation is that they are equal (since that yields the required result), which is obvious from a sketch of the graphs of cos(x) and cos([itex]\pi[/itex]/2-x). But I am not sure how you would derive this more formally?
  5. Nov 22, 2012 #4


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    By a change of variable. In the integral of f(t-x).dx, substitute x = t - u wherever x occurs (including the bounds).
  6. Nov 22, 2012 #5
    Sorry but I am still confused:

    I = int(0 to t)[f(t-x)]dx

    let u = t - x
    then du = -dx
    limits become t to 0

    I = -int(t to 0)[f(u)]du
    I = int(0 to t)[f(u)]du

    Now what? If I put x back in I just get what I started with.
  7. Nov 22, 2012 #6


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    The variable of integration is a 'dummy' variable. You can change it to anything you like so long as you do it consistently. If I = int(0 to t)[f(u)]du then it must also be true that I = int(0 to t)[f(x)]dx.
  8. Nov 23, 2012 #7
    I see, the value of the integral depends only on the limits and the form of the integrand.

    Thank you for your help haruspex and Millennial. :)
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