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Derivative and integral (confusing part)

  1. Dec 9, 2012 #1
    We know that d(cos^-1 (x/a))/dx = -1/sqrt(a^2 - x^2) (assuming a and x are positive)

    So...Why the integral of -1/sqrt(a^2 - x^2) is not equal to (cos^-1 (x/a)) + C??????????

    Instead, my teacher says it has to be -(sin^-1 (x/a)) + C because integral of 1/sqrt(a^2 - x^2) is sin^-1 (x/a) + C.....(only put the negative sign in)

    She doesn't really explain though...(when I ask)
     
    Last edited: Dec 9, 2012
  2. jcsd
  3. Dec 9, 2012 #2
    However, when I use wolmframalpha, it gives a different solution rather than the sin^-1

    http://www.wolframalpha.com/input/?i=+integrate+of+-1%2Fsqrt%28a^2+-+x^2%29+

    Here the link
     
  4. Dec 9, 2012 #3

    haruspex

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    Consider the possibility that the two solutions are really the same, only differing by a constant:wink:.
     
  5. Dec 10, 2012 #4


    Do you mean that both my and my teacher answers are right???
     
  6. Dec 10, 2012 #5

    haruspex

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    Yes. Can you see how? Hint: sin θ = cos(π/2-θ)
     
  7. Dec 11, 2012 #6

    HallsofIvy

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    Wouldn't that be terrible!:devil:
     
  8. Dec 17, 2012 #7

    Its because cos^-1(x) and sin^-1(x) are related by a constant.

    cos^-1(x) + sin^-1(x) = ∏/2

    So if you write
    ∫-1/√(a^2 - x^2) = cos^-1 (x/a) + C

    I can use the relation and write

    ∫-1/√(a^2 - x^2) = ∏/2 - sin^-1(x/a) + C

    Which is equal to

    ∫-1/√(a^2 - x^2) = - sin^-1(x/a) + C1

    Where C1 is some arbitrary constant = C + ∏/2.

    So its like haruspex said, solutions differ only by a constant.

    Hope that Helps :)
     
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