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Hidden_Twilight Group Title

anyone know how to differentiate [(a^2)(dx)]/ [x(sqrt(x^2 - a^2)]?

  • one year ago
  • one year ago

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  1. hartnn Group Title
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    there's a dx there..so, do u mean to ntegrate ?

    • one year ago
  2. hartnn Group Title
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    *integrate ?

    • one year ago
  3. Hidden_Twilight Group Title
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    yup. i tried trigonometric substitution but i would somehow end up with the answer "a" only

    • one year ago
  4. Hidden_Twilight Group Title
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    x = asecu dx = asecutanu sqrt(x^2 - a^2) = atanu therefor:

    • one year ago
  5. Hidden_Twilight Group Title
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    |dw:1372397190010:dw|

    • one year ago
  6. hartnn Group Title
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    hey thats absolutely correct! but don't forget we haven't integrated yet!

    • one year ago
  7. hartnn Group Title
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    \(\huge \int a \: du=..?\)

    • one year ago
  8. Hidden_Twilight Group Title
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    da?

    • one year ago
  9. hartnn Group Title
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    oh, and one error, \(dx = a \sec u\ tanu \: du\)

    • one year ago
  10. hartnn Group Title
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    not da, here the variable us 'u' , so 'du'

    • one year ago
  11. Hidden_Twilight Group Title
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    oh. so it becomes au?

    • one year ago
  12. hartnn Group Title
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    correct :) au +c now substitute back for 'u' x=a sec u so, u=...?

    • one year ago
  13. Hidden_Twilight Group Title
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    u = asec/x wait. does sec still have u or do i change it to x?

    • one year ago
  14. hartnn Group Title
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    u = asec/x ......no \(\large x = a \sec u \\ \large (x/a)=\sec u \\ \large u = \sec^{-1}(x/a)\) the inverse secant function!

    • one year ago
  15. hartnn Group Title
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    gotthis ?

    • one year ago
  16. Hidden_Twilight Group Title
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    ahhhh... so it becomes a(sec^-1 [x/a]) right?

    • one year ago
  17. hartnn Group Title
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    correct! don't forget the +c part :)

    • one year ago
  18. Hidden_Twilight Group Title
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    wait. does that mean i have to cancel a?

    • one year ago
  19. hartnn Group Title
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    you cannot cancel 'a' the other a is inside the inverse secant function

    • one year ago
  20. hartnn Group Title
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    a(sec^-1 [x/a]) +c will be the final answer

    • one year ago
  21. Hidden_Twilight Group Title
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    ohh...got it! thank you :)

    • one year ago
  22. hartnn Group Title
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    welcome ^_^ ask if anymore doubts...

    • one year ago
  23. Hidden_Twilight Group Title
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    btw. what if the other sie has y? a(sec^-1 [x/a]) = y (y was already integrated) a(sec^-1 [x/a]) = y + C (sec^-1 [x/a]) =( y + C)/a ?

    • one year ago
  24. hartnn Group Title
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    ohh...differential equations ? correct, go on... x/a = sec [(y+c)/a ] or x = a sec [(y+c)/a] ....

    • one year ago
  25. Hidden_Twilight Group Title
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    ahhhh....so thats how it is. thanks har!! :D

    • one year ago
  26. hartnn Group Title
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    welcome ^_^

    • one year ago
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