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anonymous

  • 5 years ago

Compute the integral of 1/(x*sqrt(x^3-25)) Hint: let u=x^(3/2)

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  1. anonymous
    • 5 years ago
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    I don't know that your hint gets you there. Who gave the hint?

  2. anonymous
    • 5 years ago
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    it is on the review sheet

  3. anonymous
    • 5 years ago
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    i know the answer too if that helps. its 2/15 arcsec(x^(3/2)/5)+c

  4. anonymous
    • 5 years ago
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    In doing it, they manipulated in a way to use trig method.

  5. anonymous
    • 5 years ago
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    Yeah, you have to brush up on your trig method (some variation of tan^2=sec2-1). Using hint x under the square root turns to sq root of u^2-25 and you can do the trig method.

  6. anonymous
    • 5 years ago
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    put (x^(3/2)-25)=t^2

  7. anonymous
    • 5 years ago
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    use this substitution

  8. anonymous
    • 5 years ago
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    I think he wants us to use the trig method. I just keep getting stuck. I have u=x^(3/2) du=3/2sqrt(x)dx 2/3du=sqrt(x)dx I would be able to solve from there but i dont have a sqrt(x)

  9. anonymous
    • 5 years ago
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    use my method and multiply denominator and numerator by x^0.5 and put x^(3/2) as t^2+25

  10. anonymous
    • 5 years ago
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    He's going to test us on the trig method...

  11. anonymous
    • 5 years ago
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    who?

  12. anonymous
    • 5 years ago
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    the teacher

  13. anonymous
    • 5 years ago
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    oh...but even ur hint uses substitution

  14. anonymous
    • 5 years ago
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    right, but isnt there a difference between trig substitution and regular substitution?

  15. anonymous
    • 5 years ago
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    if u want to use trig then use sec^(2/3)t=x

  16. anonymous
    • 5 years ago
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    got it?

  17. anonymous
    • 5 years ago
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    not really. Can you go through the process step by step?

  18. anonymous
    • 5 years ago
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    use (25*sec (t))^(4/3)=x so x^(3/2)=25*sec^2(t) so dx=25^(4/3)*(4/3)*sec^(1/3)(t)*sec(t)*tan(t)

  19. anonymous
    • 5 years ago
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    dx=c*sec^(4/3)(t)*tan(t) dt c=constant

  20. anonymous
    • 5 years ago
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    now solve further i m very slow in typing

  21. anonymous
    • 5 years ago
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    Jas, give me a minute. I have the method his instructor is expecting.

  22. anonymous
    • 5 years ago
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    OK, using hint \[u=x ^{3/2}\]Convert this all the way through you get \[u ^{2/3}=x\]

  23. anonymous
    • 5 years ago
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    if u will solve my expression then it will become simplified

  24. anonymous
    • 5 years ago
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    Yes, but he is in school, he is learning a really helpful method, technique.

  25. anonymous
    • 5 years ago
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    Just hold on one minute, let me finish, so the work don't get all muddled.

  26. anonymous
    • 5 years ago
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    i have already told similar method before but he wants trig method

  27. anonymous
    • 5 years ago
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    k......go ahead

  28. anonymous
    • 5 years ago
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    \[2/3 du =\sqrt{x}\]

  29. anonymous
    • 5 years ago
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    if u will use x^(3/2)-25=t^2 then it will be more easier

  30. anonymous
    • 5 years ago
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    New\[\int\limits_{?}^{?}[(2/3) du]/[(u ^{2/3}\sqrt{u ^{2}-25}\]

  31. anonymous
    • 5 years ago
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    what happened to 2/3du=sqrt(x)?

  32. anonymous
    • 5 years ago
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    it is 2/3du=sqrt(x)dx

  33. anonymous
    • 5 years ago
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    \[\int\limits_{?}^{?}[(2/3) du]/25\sec \theta \tan \theta\]

  34. anonymous
    • 5 years ago
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    I guess that should be derivative theta

  35. anonymous
    • 5 years ago
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    i mean, what happens to the sqrt(x)?

  36. anonymous
    • 5 years ago
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    multiply deno and nume by x and cha... made mistake in calculation

  37. anonymous
    • 5 years ago
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    In the substitution there is an extra \[\sqrt{5 \sec \theta}\] and that is = to sq rt of x. That is swallowed up in dtheta.

  38. anonymous
    • 5 years ago
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    OK, I just noticed the square root of x is in bottom. How do I get it on top to be swallowed by dtheta, Jas?

  39. anonymous
    • 5 years ago
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    you solve complete expression in u first it will be 1/u*(......)

  40. anonymous
    • 5 years ago
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    (.....)=u^2-25

  41. anonymous
    • 5 years ago
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    now put u=sec(t)

  42. anonymous
    • 5 years ago
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    One second, there is a fix for this, don't change it too much or uswag would get lost in translation.

  43. anonymous
    • 5 years ago
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    otherwise change directly which i already proposed earlier

  44. anonymous
    • 5 years ago
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    OK, that is already wiped out uswag. when I bring sq rt x and dx on top, I have an extra sq rt x in bottom, they cancel each other out.

  45. anonymous
    • 5 years ago
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    there is no sqrt but there is x

  46. anonymous
    • 5 years ago
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    Yes, but that x is converted to u^(2/3). In my conversion I keep u^2=5 sec theta and I wipe out sq root 5 theta or sq rt x.

  47. anonymous
    • 5 years ago
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    Uswag, you seem to be following along. Does this make any sense?

  48. anonymous
    • 5 years ago
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    k...i was doing in other way

  49. anonymous
    • 5 years ago
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    i need to go

  50. anonymous
    • 5 years ago
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    Can we start the explanation from 2/3 du/ u^(2/3)(sqrt(u^2-25)..... i got lost in the explanation after that. so its 2/3 du= sqrt(x)dx so we have sqrt(x)dx on top.

  51. anonymous
    • 5 years ago
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    Thanks Jas, you been a great help.

  52. anonymous
    • 5 years ago
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    Thanks Jas

  53. anonymous
    • 5 years ago
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    thanks 4 being my fan

  54. anonymous
    • 5 years ago
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    \[{[2/(3\sqrt{x})]du}/[\sqrt{x}u ^{2}\sqrt{u ^{2}-25}]\]

  55. anonymous
    • 5 years ago
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    \[2/75\int\limits_{?}^{?}d \theta/\sec \theta \tan \theta\]

  56. anonymous
    • 5 years ago
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    Now you have a function you can integrate. Once you integrate you have to go back and change to x. These trig techniques are explained nicely in MIT lectures http://www.youtube.com/watch?v=CXKoCMVqM9s&feature=BFa&list=PL590CCC2BC5AF3BC1&index=25

  57. anonymous
    • 5 years ago
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    I'll write down some of the substitutions I made\[u^{2}-25=5^{2}\sec ^{2}\theta-5^{2}\]

  58. anonymous
    • 5 years ago
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    And that is equal to \[5^{2}\tan ^{2}\theta\]

  59. anonymous
    • 5 years ago
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    \[u =5\sec \theta \]

  60. anonymous
    • 5 years ago
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    \[\sqrt{u ^{2}-25}=\sqrt{5^{2}\sec ^{2}\theta-5^{2}}=\sqrt{5^{2}\tan ^{2}\theta}=5\tan \theta\]

  61. anonymous
    • 5 years ago
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    A lot of info, makes sense?

  62. anonymous
    • 5 years ago
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    yeah, after seeing the info plus that it does. Thanks!

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