1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trigonometric Integration / relationship between f(cosx) and f(sinx)
  1. Is f integrable? (Replies: 8)

Loading...