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Changing limits of integration in subsitution

  1. Apr 8, 2014 #1

    I'm doing this integration:

    I = ∫[itex]^{1}_{-1}[/itex]1/([itex]\pi[/itex]√1-x2)dx

    I made the substitution x = cosθ, and I'm fine with performing the integral apart from changing the limits - for x = 1 I put θ = 0, but for x = -1, how do I know whether to choose θ = [itex]\pi[/itex] or θ = -[itex]\pi[/itex]? The first choice gives I = 1 but the second gives I = -1. I know arccos is defined so that 0 ≤ θ ≤ [itex]\pi[/itex], but surely choosing θ = -[itex]\pi[/itex] should work and give the same value for the integral anyway, since this is just to avoid arccos being multi valued?

  2. jcsd
  3. Apr 8, 2014 #2


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    The endpoints ##\theta = \pi## and ##\theta = -\pi## may correspond go the same point on the unit circle, but you are integrating over an entire interval, not just at that point. The intervals ##[0,\pi]## and ##[0,-\pi]## do not correspond to the same points on the circle. The first interval covers the upper half of the circle, traveling counterclockwise, and the second interval covers the lower half of the circle, traveling clockwise. You need to work out which of these intervals corresponds to your situation.

    In particular, if ##x = \cos(\theta)##, then ##\sqrt{1 - x^2} = \sqrt{\sin^2(\theta)} = |\sin(\theta)|##. If you want to write this without absolute values, then that forces a particular choice for your integration interval.
    Last edited: Apr 8, 2014
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