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anonymous

  • 5 years ago

find the intergral of 1/((1+9(x^2))^2)

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  1. anonymous
    • 5 years ago
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    What have you tried?

  2. anonymous
    • 5 years ago
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    factor out the 9, youre going to get arctan something

  3. anonymous
    • 5 years ago
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    alright i know what i need to do i need to use trig subsitution and i wasn't sure 1/((1+9(x^2))^4/2) or should i just use tan i mean what is going to be the co-effiecinet but that seems incorrect

  4. anonymous
    • 5 years ago
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    oh nevermind, i didnt see the square . you can do trig substitution

  5. anonymous
    • 5 years ago
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    ohk i can factor out the 9 that seems a good think thanks i just wasn't doing that :| i guess

  6. anonymous
    • 5 years ago
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    whoa, where did 4/2 come from?

  7. anonymous
    • 5 years ago
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    4/2 is the same thing as 2 so when i see all the examples in book they have roots under the section of trig subsitution

  8. anonymous
    • 5 years ago
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    dont use 4/2, thats weird

  9. anonymous
    • 5 years ago
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    It is weird. Very weird. Use a tan sub, as above.

  10. anonymous
    • 5 years ago
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    if youre using an integral table look for 1 / ( x^2 + a^2)^2 dx

  11. anonymous
    • 5 years ago
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    yup i am using a tan sub just sure what to put as a coefficient for tan

  12. anonymous
    • 5 years ago
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    Integral tables are for babies.

  13. anonymous
    • 5 years ago
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    yes that i what i am following cantroset but in that the a will 1/3

  14. anonymous
    • 5 years ago
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    errr you can have 1 / ( ax^2 + b^2) ^2 dx,

  15. anonymous
    • 5 years ago
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    i know that formula :P

  16. anonymous
    • 5 years ago
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    whats your table say

  17. anonymous
    • 5 years ago
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    :) so 1/3 should be my coefficient right :)

  18. anonymous
    • 5 years ago
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    lets use trig sub just to be thorough, i use pauls online math course, here one sec i will post link

  19. anonymous
    • 5 years ago
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    \[\text{Let } x = \frac{1}{3}\cdot \tan \alpha\]

  20. anonymous
    • 5 years ago
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    Paul is the man!! http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx scroll down

  21. anonymous
    • 5 years ago
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    yes why is that a 1/3 that is what i am lost about shouldn't it be 1/9

  22. anonymous
    • 5 years ago
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    ok one sec, i hate this thing it keeps freezing. yeesh why doesnt google help with the appearance of this

  23. anonymous
    • 5 years ago
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    the 1/3 is squared to 1/9

  24. anonymous
    • 5 years ago
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    ohk so it is (1+ 1/3 tan (theta))^2

  25. anonymous
    • 5 years ago
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    (1+ 1/3 tan (theta)^2)^2

  26. anonymous
    • 5 years ago
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    oh i think i see what youre doing, youre forcing it to be a square root ok sqrt a^2 + b^2x^2 , let x = a/b tan t

  27. anonymous
    • 5 years ago
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    \[1+9x^2 = 1+ 9\left(\frac{1}{3}\tan \alpha \right)^2 = 1+\tan^2 \alpha\]

  28. anonymous
    • 5 years ago
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    so you want ( 1 + 9x^2)^2 = [ (1+9x^2)^4] ^1/2 ?

  29. anonymous
    • 5 years ago
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    i think newton thats simpler,

  30. anonymous
    • 5 years ago
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    paul seems to say you want a square root though

  31. anonymous
    • 5 years ago
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    oh that makes so much more sense :)

  32. anonymous
    • 5 years ago
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    you don't really need to have it i guess you can force it but you don't need it you need it for inital i guess it won't really change anything ( i am referring to the square root)

  33. anonymous
    • 5 years ago
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    xD how on earth does that make more sense. Crazy Americans.

  34. anonymous
    • 5 years ago
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    you can also do partial fractions i bet

  35. anonymous
    • 5 years ago
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    i mean what you wrote.

  36. anonymous
    • 5 years ago
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    yeah i can do partial fraction, but i am required to do trig :)

  37. anonymous
    • 5 years ago
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    Oh, good. And you can't split it in to partial fractions (I don't think).

  38. anonymous
    • 5 years ago
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    here integrals involving quadratics http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithQuadratics.aspx

  39. anonymous
    • 5 years ago
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    why not?

  40. anonymous
    • 5 years ago
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    Because it's the same term squared on the bottom and no others.

  41. anonymous
    • 5 years ago
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    But assuming you could put it into partial fractions, it would not simplify it at all.

  42. anonymous
    • 5 years ago
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    so those are your factors irreducible quadratic

  43. anonymous
    • 5 years ago
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    (9x^2+1)^2 = (tan(a)^2+1) = sec(a)^4 is this correct

  44. anonymous
    • 5 years ago
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    but probably not the easiest approach. youre tan sub is fine , one sec

  45. anonymous
    • 5 years ago
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    No, because it would be \[\frac{ax + b}{1+9x^2} + \frac{cx + d}{\left(1+9x^2\right)^2}\] But they all equal 0 except fro d = 1, because ... etc

  46. anonymous
    • 5 years ago
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    avnis yes

  47. anonymous
    • 5 years ago
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    Now include the change of what you are integrating with respect to and go from there.

  48. anonymous
    • 5 years ago
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    ohk so next 1/sec(a)^4 = cos(a)^4

  49. anonymous
    • 5 years ago
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    Yes, but\[\frac{\mathbb{d}\alpha}{\mathbb{d}x}\] don't forget to include

  50. anonymous
    • 5 years ago
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    not if you use complex root factoring

  51. anonymous
    • 5 years ago
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    what

  52. anonymous
    • 5 years ago
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    avnis just talking to newton

  53. anonymous
    • 5 years ago
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    Oh, OK, you can;t split it up into partial fractions in a way which would make it a retardedly different problem - is that better?

  54. anonymous
    • 5 years ago
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    newton, it would be like partial decomp'ing 1/(x^2 + 1) = A/ (x+i) + B/ ( x -i)

  55. anonymous
    • 5 years ago
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    i prefer approaches , one size fits all. hehe, the one size hammer that is

  56. anonymous
    • 5 years ago
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    how do i solve the rest of it

  57. anonymous
    • 5 years ago
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    @anvis If you are now integrating with respect something other than x, you need to account for that. Note: \[x = \frac{1}{3}\tan\alpha \implies \frac{\mathbb{d}x}{\mathbb{d}\alpha} = \frac{1}{3}\sec^2 \alpha\]

  58. anonymous
    • 5 years ago
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    i see which will make a equal to tan^-1(3x)

  59. anonymous
    • 5 years ago
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    so i wll be left with 1/3 integral of cos^2(a)

  60. anonymous
    • 5 years ago
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    It will, but you don't need to include that yet. \[\int\limits \frac{1}{(1+9x^2)^2} \mathbb{d}x\] becomes \[\int\limits \frac{1}{(1+\tan^2\alpha)^2} \cdot \frac{1}{3} \sec^2 \alpha \cdot \mathbb{d}a = \int \frac{1}{3} \cos^2\alpha\ \cdot \mathbb{d}a\] yes.

  61. anonymous
    • 5 years ago
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    Now do whatever you feel like with cos^2 to make it easier to integrate, sub back in x with the tan^(-1) thing you had above and you're done.

  62. anonymous
    • 5 years ago
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    1/3inetgarl (1/2 (cos 2(a)+1/2) du

  63. anonymous
    • 5 years ago
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    great job

  64. anonymous
    • 5 years ago
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    you have an extra 1/2 there

  65. anonymous
    • 5 years ago
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    oh yes

  66. anonymous
    • 5 years ago
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    Sorry gang, got to go. But here's a nice one for you to try if you feel up to it: By finding constants a, b, c and d such that: \[\frac{ax+b}{x^2+2x+2} + \frac{cx+d}{x^2-2x+2} \equiv \frac{1}{x^4+4} \] Show \[\int_0^1\frac{1}{x^4+4} = \frac{1}{16}\ln 5 + \frac{1}{8}\tan^{-1}2\]

  67. anonymous
    • 5 years ago
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    ill solve it

  68. anonymous
    • 5 years ago
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    hmm so i got a/6 + (sin 2a)+12 and i know that a= tan^-1(3x) how i do i solve that

  69. anonymous
    • 5 years ago
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    you make a triangle

  70. anonymous
    • 5 years ago
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    what would the side be would one side be 3x

  71. anonymous
    • 5 years ago
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    ...

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  72. anonymous
    • 5 years ago
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    that is what i got

  73. anonymous
    • 5 years ago
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    but there is this online answer from wolfaram and it doesn't match it so i am so lost

  74. anonymous
    • 5 years ago
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    What did you get as sin(2x) ?

  75. anonymous
    • 5 years ago
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    2sin(x)cos(x)

  76. anonymous
    • 5 years ago
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    OK, what what did you get as sin(x) and cos(x)? Hint, use the trinagle I uploaded, with the facts opp/hyp = sin(x) and adj/hpy = cos(x) NOTE these should all be as, NOT xs, sorry for the confusion.

  77. anonymous
    • 5 years ago
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    ya that is why i was thinking i don't have x:|

  78. anonymous
    • 5 years ago
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    sin(a) =3x/ sqrt (9x^2+1)

  79. anonymous
    • 5 years ago
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    Yeah the typing on my diagram should be a, too. Indeed. And cos(a)? And then 2sin(2a) ?

  80. anonymous
    • 5 years ago
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    sorry, only sin(2a), not 2 sin(2a)

  81. anonymous
    • 5 years ago
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    cos (a)= 1/ sqrt(9x^2+1)

  82. anonymous
    • 5 years ago
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    :D

  83. anonymous
    • 5 years ago
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    got it :)

  84. anonymous
    • 5 years ago
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    Woo

  85. anonymous
    • 5 years ago
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    i have learnt alot today :)

  86. anonymous
    • 5 years ago
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    Good to hear. Goodbye.

  87. anonymous
    • 5 years ago
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    thanks :)

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