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Solving integral from 5/(x^2 + 1) dx from -1 to 1 works one way but

  1. Sep 16, 2013 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The integral of 5/(x^2 + 1) dx from -1 to 1. (The TheIntegral.jpg attachment shows this in a aesthetically-pleasing way.)

    2. Relevant equations
    sin^2 (θ) + cos^2 (θ) = 1
    1 + tan^2 (θ) = sec^2 (θ)
    cot^2 (θ) + 1 = csc^2 (θ)

    x = tan(θ)
    x = cot(θ)

    3. The attempt at a solution
    When I am solving the integral, I successfully compute it using the x = tan(θ) substitution but when I use the x = cot(θ) substitution, I get the wrong answer.

    My specific work is attached as MyWork.jpg. The work in black is the work I got correct (as indicated by the red writing) and, the work in green is the work I get wrong (as indicated by the red writing).

    Could someone please tell me what I am doing wrong when using the cot(θ) substitution?
     

    Attached Files:

  2. jcsd
  3. Sep 16, 2013 #2
    [itex]\text{cot}^{-1}1\neq \frac{1}{\tan^{-1}1}[/itex]

    The inverse trig functions do not satisfy the same identities as the trig functions.
     
  4. Sep 16, 2013 #3

    pasmith

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    [itex]\cot(\theta)[/itex] blows up when [itex]\tan(\theta) = 0[/itex], ie. when [itex]\theta = n\pi[/itex].

    So you need to find values for [itex]\cot^{-1}(\pm 1)[/itex] which lie between [itex]0[/itex] and [itex]\pi[/itex]. [itex]\cot^{-1}(1) = \frac{\pi}{4}[/itex] and [itex]\cot^{-1}(-1) = \frac{3\pi}{4}[/itex] will do.
     
  5. Sep 16, 2013 #4

    s3a

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    Thanks, gother_p, for specifying.

    Also, pasmith, thanks for your answer too however, although I can see how using 3π/4 would work algebraically, I don't get how I would know to choose 3π/4 over –π/2; when θ = –π/2, θ ≠ nπ (for n being any integer).

    Also, instead of values between 0 and π, I could choose values between 0 and –π if I wanted, right?
     
  6. Sep 17, 2013 #5

    s3a

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    Actually, I meant θ = –π/4 instead of θ = –π/2.

    Also, I'm still confused as to why I can't choose θ = –π/4 instead of θ = 3π/4 (for the reason I listed in my previous post) and, I would really appreciate it if this can be elaborated upon.

    Lastly, if I'm not being clear, tell me what to clarify and, I will do so.
     
  7. Sep 18, 2013 #6

    eumyang

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    You can't choose [itex]\theta = -\frac{\pi}{4}[/itex], because by definition the range of the inverse cotangent function [itex]\theta = \cot^{-1} x[/itex] is [itex]0 < \theta < \pi[/itex]. There is only one value for [itex]\cot^{-1} \left( -1 \right)[/itex] and that is [itex]\frac{3\pi}{4}[/itex].
     
  8. Sep 18, 2013 #7

    s3a

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    Oh, I think I get it now!

    To attempt to restate what you guys said using graphical terminology, is it because in the graph of f(x) = cot(x), there is an asymptote separating –π/4 and π/4 so, we could choose to replace –π/4 with 3π/4, for example, because, the cotangent of –π/4 and 3π/4 yield the the same value or, equivalently, we could choose any other two angles both of which are between the same asymptotes such as –π/4 and –3π/4?
     
  9. Sep 18, 2013 #8

    pasmith

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    Yes.
     
  10. Sep 18, 2013 #9

    s3a

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    Thanks for confirming. :)
     
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