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Homework Help: U-substitution - prove .

  1. Oct 6, 2009 #1
    help with u-substitution - prove.....

    1. The problem statement, all variables and given/known data
    Some function f is continuous on [0,Pi]

    Prove: [tex]\int_0^{\pi}\\xf(sin x)\,dx = \frac{\pi}{2}\right)\int_0^{\pi}\\f(sinx)dx[/tex]

    using the substitution [tex]u=\pi-x[/tex].

    2. Relevant equations
    Identities, integral properties.

    3. The attempt at a solution
    I've tried this two ways, (1) trying to transform the integral on the left ([tex]\int_0^{\pi}\\xf(sin x)\,dx[/tex]) to the one on the right and (2) vice versa.

    1) If I am to use the u-sub: [tex]u=\Pi-x[/tex], then:

    [tex]du=d(\Pi-x)=-dx[/tex] <==> [tex]-du=dx[/tex]

    First, from the identity:


    I re-expressed the integral like such:


    Then substituted the quantity u into the integrand:




    Now here's where I get stuck, after much trial and error, ultimately leading to no where, I've found that [tex]\frac{d}{dx}\right)(\frac{F(x^2)}{2}\right)) = xf(x^2)[/tex].

    2) From the left side of the statement.


    When I start from the side, it appears as though the quantity [tex]u = \frac{2}{\Pi}\right)x[/tex] and the differential [tex]du=\frac{2}{\Pi}\right)dx[/tex] but then obviously I can't set u=Pi-x.

    Last edited: Oct 6, 2009
  2. jcsd
  3. Oct 7, 2009 #2


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    Gold Member

    Re: help with u-substitution - prove.....

    I'm not sure what you are doing here. This has taken you right back where you started.

    If [itex]u = \pi - x[/itex] then [itex]du = -dx[/itex] as you indicated, so

    [tex]\int_0^\pi x f(\sin x) dx = \int_0^\pi (\pi - u) f(\sin(\pi - u)) du =
    \int_0^\pi (\pi - u) f(\sin u) du[/tex]

    Note that there's nothing special about the letter [itex]u[/itex] on the right hand side. You can just as well replace it with [itex]x[/itex]. If you do that, the answer is actually right in front of your eyes, in a slightly tricky form.
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