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Derivation of some integral formulas

  1. Mar 14, 2015 #1
    Hi,

    The integrals in the image are well-known, but we are not allowed to use them in the test.

    I have tried to prove them using changing variable t, but I didn't succeed.

    Can you show me the way? YvnnR3H.jpg
     
  2. jcsd
  3. Mar 14, 2015 #2

    Ray Vickson

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    Show us your actual work on at least one example above---whether it failed or not.
     
  4. Mar 14, 2015 #3

    HallsofIvy

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    What do you mean by "not allowed to use them"? How would being able to prove them help you on a test?

    The first and fourth can be done by considering the derivatives of arctan and arcsin. The second and third by factoring and "partial fractions".
     
  5. Mar 14, 2015 #4

    SammyS

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    For the first integral, try the trig substitution, ## x = a\cdot\tan(\theta)\ ##.
     

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  6. Mar 14, 2015 #5
    Geez I didn't think it was necessary to use a trig substitution for any of these. In the first three you can just multiply the numerator & denominator by 1/a2 & in the last two multiply numerator & denominator by 1/a.
     
  7. Mar 14, 2015 #6
    ok,

    I succeed proving number one and four- please check if I did it correctly.
    But I still didn't understand how to prove the rest... how do I use partial fractions?
    can you show me the way on one of the other integrals? 26R3FQy.jpg tjaQ3Wg.jpg
     
  8. Mar 15, 2015 #7
    I have tried to do that in your way in the second integral and didn't succeed.
    Can you show me your way?
     
  9. Mar 16, 2015 #8
    Hey, I'm on my mobile right now so I can't see your proves for first and four integrals but for the second and third use simple fractions. I mean, you have the denominator

    a^2 - x^2 = (a+x)(a-x)

    So you can write

    1/(a^2 - x^2) = 1/((a+x)(a-x)) = A/(a+x) + B/(a-x)

    For some constants A, B which you should find easily then you just have a sum of two integrals which are two logarithms and using a basic property of log you will get what you are looking for. Third integral is analogous to this.

    Should I said why take abs value?
     
  10. Mar 17, 2015 #9
    Tnx, now I got it.

    But I still didn't understand how to solve the last integral...

    I need to use some substitution t?
     
  11. Mar 17, 2015 #10
    I think in the last one you need to use the substitution x = a⋅sec θ then dx = a⋅secθ⋅tanθ dθ etc
     
  12. Mar 17, 2015 #11

    SammyS

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    That works for the " - " sign. Alternatively, you can use x = a ⋅ cosh(u) .

    Use x = a⋅sinh(u) for the "+" sign.

    Of course the results will look a bit different: cosh-1(x/a) & sinh-1(x/a) .
     
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