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Integration by trigonometric substitution

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data

    ∫ 1/(x√(25-x^2)) dx

    2. Relevant equations

    by trigonometic substitution I got the following equations
    x = 5 sin θ
    dx = 5 cos θ dθ
    √(25-x^2) = 5 cos θ
    θ = sin^-1 (x/5)

    3. The attempt at a solution

    I solved it and got the following (so far, it is okay with wolfram alpha):

    ∫ 1/(x√(25-x^2)) dx = (1/5) ∫ csc θ dθ

    according to me, the result should be (1/5) ( ln ( 5/x - √(25-x^2)/x ) )
    but according to wolfram alpha it should be (1/5) ( ln (x) - ln ( √(25-x^2) + 5 ) )

    are these two results equivalent?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 17, 2011 #2

    LCKurtz

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    Differentiate the answers and see if you get the integrand. That will tell you.
     
  4. Oct 17, 2011 #3

    SammyS

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    [itex]\displaystyle \ln (x) - \ln ( \sqrt{25-x^2} + 5 )=\ln\left(\frac{x}{\sqrt{25-x^2} + 5}\right)[/itex]
    [itex]\displaystyle =-\ln\left(\frac{\sqrt{25-x^2}}{x}+\frac{5}{x}\right)[/itex]​

    So, there's only a disagreement about the sign.
     
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