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Integration of 1/(x^2 + a^2)

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  1. Feb 23, 2016 #1
    image.jpg
    I cannot understand the intergration done here
    The part how 1/a came, what happened to the x and how did tan come in to this
     
  2. jcsd
  3. Feb 23, 2016 #2

    Samy_A

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    The x went away because it is the dummy integration variable in a definite integral.

    For starters: do you know how to evaluate the following indefinite integral: ##\int \frac{1}{1+x²}dx##?
     
  4. Feb 23, 2016 #3
    No
     
  5. Feb 23, 2016 #4

    Samy_A

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    Do you know how to use a substitution in order to compute an integral?
     
  6. Feb 23, 2016 #5
    Yes
     
  7. Feb 23, 2016 #6

    Samy_A

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    Fine. So start with the indefinite integral ##\int \frac{1}{1+x²}dx## and use the substitution ##x=\tan y## to compute it.
     
  8. Feb 23, 2016 #7
    Thank you very much
     
  9. Feb 23, 2016 #8
    image.jpg
    This was as far as I could go
    I'm wondering how that 1/a came and how to make this into a 1/x^2+1 formate so I can input tan
    Please help
     
  10. Feb 23, 2016 #9

    Samy_A

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    You can do it in two (very similar) ways.
    I assume that you found the indefinite integral ##\int \frac{1}{1+x²}dx##.
    To calculate the indefinite integral ##\int \frac{1}{a²+x²}dx##, you could:
    1) use the substitution ##x=a \tan y## and solve the same way as you did for ##\int \frac{1}{1+x²}dx##;
    2) use the substitution ##x=ay##, which gives ##\int \frac{1}{a²+x²}dx =\int \frac{1}{a²+a²y²} ady =\frac{1}{a} \int \frac{1}{1+y²} dy##, the indefinite integral you already solved (up to a constant 1/a).

    Just to be clear, all my integrals here are indefinite integrals. When you calculate your definite integral, watch the integration limits when you perform a substitution.
     
  11. Feb 23, 2016 #10
    I finally figured it out image.jpg
    It can be taken like this
    It's sort of like the perfect square rule this once I put it to this formate it's done
    Thank you very much
     
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