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Homework Help: Having problems evaluating definite integral

  1. Feb 9, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm trying to evaluate
    [tex]\int_0^{2\pi} \frac{9-6\cos t}{5-4\cos t} dt \ .[/tex]
    I found that the antiderivative of the integrand is
    [tex]F(x) = \frac{3}{2}x + \tan^{-1} \left(3\tan\frac{x}{2} \right)[/tex]
    so using the FTC, the integral should be
    [tex]F(2\pi)-F(0)=3\pi[/tex]
    But using a calculator to evaluate the integral yields [itex]4\pi[/itex].
    What am I doing wrong here?

    2. Relevant equations

    3. The attempt at a solution
     
  2. jcsd
  3. Feb 9, 2012 #2
    Hint: tan-1 0 is not always equal to 0.
     
  4. Feb 9, 2012 #3
    I don't understand what you mean. If tan-1 x were multivalued, then it would not be a function, so I implicitly restricted tan-1 x to its principal branch.
     
  5. Feb 9, 2012 #4

    Ray Vickson

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    You need to worry about whether or not the hypotheses of the fundamental theorem of calculus hold; that is, whether or not [itex] \int_0^{2 \pi} f(x) \, dx = F(2\pi) - F(0) [/itex] actually holds (where F is an antiderivative of f). If F(x) sweeps over points arising from different branches of 'arctan' as x goes from 0 to 2π, you might run into trouble. To be safe, break up the integral as [itex]I = I_1 + I_2,[/itex] where x goes from 0 to π in the first integral and from π to 2π in the second; then change variables to x = π + y in the second integral, so you again have an integral from 0 to π.

    RGV
     
  6. Feb 9, 2012 #5
    Aren't the hypotheses of the FTC just that F is differentiable and F'=f? These seem to hold for the function I gave, so I don't see exactly where the problem is.
     
  7. Feb 9, 2012 #6

    SammyS

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    Yes, you need F to be differentiable.

    If [itex]\displaystyle F(x) = \frac{3}{2}x + \tan^{-1} \left(3\tan\frac{x}{2} \right)\,,[/itex] notice that F has jump discontinuities at x = (2k+1)π. In other words, F(x) is not differentiable when x is an odd integer multiple of π.

    This requires you to break up the integral as has been suggested.
     
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