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SABREEN

how do we integrate z/(x^2+z^2) with respect to x the answer is arctan(x/z)

  • one year ago
  • one year ago

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  1. VeritasVosLiberabit
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    \[\int\limits_{}^{}\frac{ z }{ x ^{2}+z ^{2} }dx\]

    • one year ago
  2. VeritasVosLiberabit
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    this is a formula you use to compute the integral of a function with this structure.

    • one year ago
  3. Rosh007
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    You will be able to relate your question simply to the standard integral \[\large \frac{1 }{ a^2+x^2 } =\frac{1 }{a } \times \arctan (\frac{x}{ a })\]

    • one year ago
  4. Rosh007
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    if Left hand side is the function that has to integrate with respect to x

    • one year ago
  5. SABREEN
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    is 1/a the derivative of x/a

    • one year ago
  6. SABREEN
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    is tht why its there?

    • one year ago
  7. VeritasVosLiberabit
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    1/a is part of the standard formula. I'll work it out for you.

    • one year ago
  8. VeritasVosLiberabit
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    \[z \int\limits_{}^{}\frac{ 1 }{ x ^{2}+z ^{2} }dx \]

    • one year ago
  9. SABREEN
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    thnk you id like to see why

    • one year ago
  10. VeritasVosLiberabit
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    \[z*\frac{ 1 }{ z }\arctan(\frac{ x }{ z })\rightarrow \arctan(\frac{ x }{ z })\]

    • one year ago
  11. VeritasVosLiberabit
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    in this case z is a scalar.

    • one year ago
  12. VeritasVosLiberabit
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    so typically z would be represented as a number. I think they are using zahlen which denotes integer values.

    • one year ago
  13. SABREEN
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    why is arc tan (x/z) and where does the 1/z come from??

    • one year ago
  14. VeritasVosLiberabit
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    1/z is always part of the equation. if you want me to do a proof, I'm afraid I don't have enough time for that but maybe someone else would.

    • one year ago
  15. Rosh007
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    the formula for most of the integral is coming fromits derivative remember derivation is the opposite of integration and wise versa. can you say what is th derivative of f(x) = arctan x ??

    • one year ago
  16. SABREEN
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    it is 1/1+x^2

    • one year ago
  17. Rosh007
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    well,good.. then what will be derivative of arctan(x/a)

    • one year ago
  18. SABREEN
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    i dunno

    • one year ago
  19. Rosh007
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    ok.. instead of x it is x/a.. substitute that value in your answer above

    • one year ago
  20. Rosh007
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    and rember x/a is another function so you need to apply the chain rule of differentiation

    • one year ago
  21. SABREEN
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    1/(1+(x/a)^2)*(1/a)

    • one year ago
  22. Rosh007
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    Yup........... You got it.. simplify it

    • one year ago
  23. malevolence19
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    I mean, I can do the integral if you want it in all the gruesome detail...

    • one year ago
  24. SABREEN
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    a/(x2+a^2)

    • one year ago
  25. SABREEN
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    yea shure if its not too complicated

    • one year ago
  26. Rosh007
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    well you are right ... now compare with your question......you got your answer

    • one year ago
  27. Rosh007
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    that is you got \[\frac{ d }{ dx }(\arctan(x/a) = \frac{ a }{ x^2+a^2 }\]

    • one year ago
  28. malevolence19
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    I mean, there isn't much too it though. The formula is basically where it comes from... Now that I think about it, I don't know how much you can actually write. I was running it by wolfram and all they do is a variable substitution and then use that int (1/u^2+1)=arctan(u) :/

    • one year ago
  29. Rosh007
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    now itegrat on both sides.. you got answer

    • one year ago
  30. Rosh007
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    Remember Integration is the opposite of differentiation @SABREEN .. now what will be tha answer?

    • one year ago
  31. SABREEN
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    so than the integral of a/(x^2+a^2) is arctan (x/a)

    • one year ago
  32. SABREEN
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    thanks=)

    • one year ago
  33. Rosh007
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    yes.. Welcome.. Now should I explain how 1/a comes ???? ;)

    • one year ago
  34. malevolence19
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    Next one in line: \[\int\limits \tanh^{-1}(ax)dx\]

    • one year ago
  35. SABREEN
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    loll nahh i got this bro=)

    • one year ago
  36. Rosh007
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    @SABREEN ....Good... Welcome and enjoy.... @malevolence19 .......you could solve by following the method i described here...only difference is x = ax and remember ax is a fuction.. :)

    • one year ago
  37. malevolence19
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    I mean, I know how to to do it lol. It's a integration by parts problem once you look up the derivative to tanh^(-1)(x)

    • one year ago
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