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experimentX

  • 2 years ago

Just a similar Q. Integrate with subs:- \[ 1. \int \sqrt{(x-a)(x-b)} \; dx\] \[ \cancel{ 2.\int \sqrt{(x^2-a)(x^2-b)} \; dx}\] don't try .. 2. does is not closed in elementary functions.

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  1. mukushla
    • 2 years ago
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    watching :)

  2. experimentX
    • 2 years ago
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    hahah ....

  3. TuringTest
    • 2 years ago
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    I think I'm just gonna watch and learn as I'm trying to eat breakfast and this technique is brand new to me :)

  4. experimentX
    • 2 years ago
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    me too ... just for the sake of memory :)

  5. Callisto
    • 2 years ago
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    Me three!

  6. TuringTest
    • 2 years ago
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    This is why OS is awesome, they never teach you subs like this in the USA...

  7. experimentX
    • 2 years ago
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    there seems to be a substitution similar to Weierstrass subs ... called Euler's subs (not surprised to hear Euler) http://en.wikipedia.org/wiki/Euler_substitution let's see if we can find something any better.

  8. experimentX
    • 2 years ago
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    this looks promising \[ x = a \; \sec ^2 t - b\; \tan ^2 t\]

  9. experimentX
    • 2 years ago
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    looks like Euler's subs is like Weierstrass subs ... always gives the answer but really last resort.

  10. klimenkov
    • 2 years ago
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    Can I try the first one?

  11. experimentX
    • 2 years ago
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    yep sure ...

  12. klimenkov
    • 2 years ago
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    1. Using the sub \(x=t+\frac{a+b}2\), so \(\int \sqrt{(x-a)(x-b)} \; dx=\int \sqrt{(t-\frac{a-b}2)(t+\frac{a-b}2)}\;dt=\int\sqrt{t^2-\frac{(a-b)^2}4}\;dt\) Now let \(\frac{a-b}2=c\) use \(t=c\cosh u,\; dt=c\sinh udu\): \(\int\sqrt{t^2-c^2}\;dt=\int c^2\sinh^2u\;du =\frac{c^2}{2}(\frac12\sinh2u-u)+C\) Now I must think how to make an inverse substitution...

  13. experimentX
    • 2 years ago
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    pretty cool technique ... i hadn't seen this either.

  14. Zekarias
    • 2 years ago
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    you see how you change dx?

  15. experimentX
    • 2 years ago
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    dx = dt

  16. Zekarias
    • 2 years ago
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    I get it.

  17. experimentX
    • 2 years ago
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    nice use of symmetry on the previous solution BTW

  18. klimenkov
    • 2 years ago
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    But there is a question how to find the real solution? How \(\sinh 2u\) can be found from \(t=\cosh u\)?

  19. experimentX
    • 2 years ago
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    that is just inverse of cosine hyperbolic ...

  20. experimentX
    • 2 years ago
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    \[ \cosh^{-1}\left( t \over c \right) = u \]

  21. experimentX
    • 2 years ago
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    this should be equal to \[ \log \left( {t \over c} + \sqrt{\left( t \over c\right)^2 - 1 } \right ) \]

  22. klimenkov
    • 2 years ago
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    Ok. But I don't like \(\sinh(2\;\text{arccosh}\; t) \)

  23. experimentX
    • 2 years ago
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    i don't like hyperbolic functions either. because they look like circle in complex plane :)

  24. klimenkov
    • 2 years ago
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    But this will make easier. Euler subs will make the integral too complicated.

  25. experimentX
    • 2 years ago
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    yeah ... agree on that point. I don't like Weierstrass subs either.

  26. klimenkov
    • 2 years ago
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    I calculated this in Wolfram Mathematica and found that it is not so similar as you have said in the beginning!

  27. klimenkov
    • 2 years ago
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    What is your method to find the first integral? Do you have any ideas how to calculate the second one?

  28. experimentX
    • 2 years ago
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    i would try makings \[ x = a \; \sec ^2 t - b\; \tan ^2 t \] haven't thought of second one.

  29. klimenkov
    • 2 years ago
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    Open this link and cry. http://www.wolframalpha.com/input/?i=Integrate[Sqrt[%28x^2+-+a%29*%28x^2+-+b%29]%2C+x]

  30. experimentX
    • 2 years ago
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    something is really bad with chrome

  31. experimentX
    • 2 years ago
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    damn ... lol. Didn't know it would get involved with two elliptical types.

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