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anonymous

  • 5 years ago

find dy/dx for y=integral(arctan)tdt from (1/x,4x)

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  1. amistre64
    • 5 years ago
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    \[y = \int\limits_{} \tan^{-1}(t) dt\] is this the equation?

  2. anonymous
    • 5 years ago
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    yes but it is a definite integral 1/x to 4x

  3. amistre64
    • 5 years ago
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    we might have to convert this to a trig of "u"... maybe. I dont know any derivatives that become an arctan...

  4. amistre64
    • 5 years ago
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    if we let u = tan^-1(t) then; t = tan(u) ; dt = sec^2(u) du which gives us... [S] u sec^2(u) du ; to solve

  5. amistre64
    • 5 years ago
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    does that make sense so far?

  6. amistre64
    • 5 years ago
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    i wonder if we can decompose this......

  7. amistre64
    • 5 years ago
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    I dont see any easy ways of doing this :)

  8. anonymous
    • 5 years ago
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    i think i'm just making it more difficult than it needs to be in some way

  9. anonymous
    • 5 years ago
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    do you understand piecewise functions?

  10. amistre64
    • 5 years ago
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    is there an answer that you can chaeck against?

  11. amistre64
    • 5 years ago
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    i know if piecewise functions; but havenet come across a way to integrate them

  12. anonymous
    • 5 years ago
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    mmk and there isn't an answer to check :/ thanks for trying though!

  13. anonymous
    • 5 years ago
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    .-. I don't think that's right amistre though I didn't try it myself lol >_<

  14. anonymous
    • 5 years ago
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    wh are you letting tan^-1(t) = u? ._.

  15. amistre64
    • 5 years ago
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    this is prolly an integration by parts method; but I dont know much about when it goes into multiple integration by parts :)

  16. amistre64
    • 5 years ago
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    becasue there is no function that derives down to a tan^-1

  17. amistre64
    • 5 years ago
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    it is equivalent to: [S] u sec^2(u) du tho.... and thats doable somehow

  18. anonymous
    • 5 years ago
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    integration by parts is something like this:\[\int\limits f'gdx = fg - \int\limits fg'dx\] where f' = sec^2(u) and g = u then find f and g' and plug it in the equation ^_^

  19. anonymous
    • 5 years ago
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    lol I got you amistre :)

  20. amistre64
    • 5 years ago
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    int by parts is still a mystery to me for some functions :)

  21. anonymous
    • 5 years ago
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    it's alright ^_^ all you have to do is find f and g' and then plug it in the equation, with that, you have solved the problem. g' = 1 and f = tan x then you'll get : \[\int\limits f'gdx = (\sec^2(u))(u) - \int\limits \tan(u) du\] then just find the integral lol

  22. amistre64
    • 5 years ago
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    i usually do a table like this: | v up ----------- u dn | sec^2 ----------- +u | tan ----------- -1 | ?? ---------- +0 | N/A ---------- And so I get for the integral: u tan(u) - 1(????)

  23. anonymous
    • 5 years ago
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    \[\int\limits f'gdx = (\sec^2(u))(u) + \ln|\cos(u)| + c\] I hope I'm right, I just followed your given lol

  24. anonymous
    • 5 years ago
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    O_O what kind of table is THAT LOL

  25. amistre64
    • 5 years ago
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    its an integration by parts table; but it lines everything up as tho it was multiple integration by parts :)

  26. anonymous
    • 5 years ago
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    .-. why so complicated? LOL

  27. anonymous
    • 5 years ago
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    you can do it my way :) much easier ^_^

  28. amistre64
    • 5 years ago
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    its easier when doing multiple integration by parts; which is all the table amounts to. You shift between + and - your u derives down and your v suits up.... its the same process :) just a different format to keep everything in line :)

  29. anonymous
    • 5 years ago
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    >_< lol, I've learned something new today! thank you :)

  30. amistre64
    • 5 years ago
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    :) all I need is an integral for tan(u) and I got it solved lol

  31. amistre64
    • 5 years ago
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    the table of integration says: tan(x) ints to ln(sec(x)) cool :)

  32. amistre64
    • 5 years ago
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    u tan(u) - ln(sec(u))

  33. anonymous
    • 5 years ago
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    hmm: \[\tan x = \frac{sinx}{cosx}\] let u = cos x and du = -sinx and solve lol

  34. amistre64
    • 5 years ago
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    we re-stitute for u and get: tan^-1(t) tan(tan^-1(t)) - tan(tan^-1(t))

  35. anonymous
    • 5 years ago
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    .-.

  36. amistre64
    • 5 years ago
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    well...... ln(sec(tan^-1(t))) lol got excited ;)

  37. anonymous
    • 5 years ago
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    not convinced with the sec u amistre :(

  38. amistre64
    • 5 years ago
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    can always try to derive it back ;)

  39. anonymous
    • 5 years ago
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    tan x = sinx/cosx integrating that will give you ln|cosx|

  40. anonymous
    • 5 years ago
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    >_<lol

  41. anonymous
    • 5 years ago
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    \[\tan ^{-1}t=u, tanu = t\] \[dt=\sec ^{2}u du=(1 + \tan ^{2}u)d u\]

  42. anonymous
    • 5 years ago
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    ._.

  43. anonymous
    • 5 years ago
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    students have faced such a question last semester, thankfully I wasn't there lol , anyhow, thanks for your time amistre and uzma ^_^ but I gotta study communications , take care both :)

  44. anonymous
    • 5 years ago
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    :) you too

  45. anonymous
    • 5 years ago
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    :)

  46. anonymous
    • 5 years ago
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    \[\int\limits_{?}^{?}\tan ^{-1}u=\int\limits_{?}^{?}1.\tan ^{-1}u\]

  47. anonymous
    • 5 years ago
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    integrating by parts \[\tan ^{-1}u.u-\int\limits_{?}^{?}u.1/(1+u ^{2})d u\]

  48. anonymous
    • 5 years ago
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    \[utan ^{-1}u -2 \int\limits_{?}^{?}2u/(1+u ^{2})\]

  49. amistre64
    • 5 years ago
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    if I did it right; mine derives back to tan^-1(t)....

  50. anonymous
    • 5 years ago
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    \[utan ^{-1}u - 2 \ln (1+u ^{2})\]

  51. anonymous
    • 5 years ago
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    i used integration by parts method, not substitution

  52. amistre64
    • 5 years ago
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    \[D[\tan^{-1}(t).\tan(\tan^{-1}(t)) - \ln(\sec(\tan^{-1}(t)))]\] \[\frac{\tan^{-1}(t)\sec^2(\tan^{-1}(t))}{t^2 +1}+\frac{\tan(\tan^{-1}(t))}{t^2+1}- \frac{\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))}{\sec(\tan^{-1}(t)).(t^2+1)}\]

  53. amistre64
    • 5 years ago
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    \[\frac{\tan(\tan^{-1}(t)).\sec^2 (\tan^{-1}(t)).\sec(\tan^{-1}(t))+\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))-\sec(\tan^{-1}(t)).\tan(\tan^{-1}(t))}{\sec(\tan^{-1}(t)).(t^2+1)}\]

  54. amistre64
    • 5 years ago
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    all that work and it cuts it short lol...; that last part is zeros the middle part to get:\[\frac{\tan^{-1}(t).\sec^2(\tan^{-1}(t)).\sec(\tan^{-1}())}{\sec(\tan^{-1}(t)).(t^2+1)}\] had to fix a typo in that second round... it aint a tan(tan^-1(t)))

  55. anonymous
    • 5 years ago
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    so the result?

  56. amistre64
    • 5 years ago
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    \[\frac{\tan^{-1}(t).\sec(^2(\tan^{-1}(t))}{t^2+1}\]

  57. anonymous
    • 5 years ago
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    n this is what u wanted?

  58. anonymous
    • 5 years ago
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    i guess i should terminate at Tan^(-1)t?

  59. amistre64
    • 5 years ago
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    since t^2+2 = sec^2(a) = sec^2(tan^-1(t)); those cancel and we are left with.. tan^-1(t)

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  60. amistre64
    • 5 years ago
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    ...since t^2 + 1.....typoed it :)

  61. anonymous
    • 5 years ago
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    hmmmm,right :)

  62. anonymous
    • 5 years ago
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    did u check mine?is it wrong?

  63. amistre64
    • 5 years ago
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    i had enough trouble proofing mine lol...my computer is bogging with all thisequation editor stuff....lol

  64. anonymous
    • 5 years ago
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    yup...hats off to ur patience :P

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