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anonymous

  • 5 years ago

integration

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{.5}8dx/(4x^2+1)^2\]

  2. amistre64
    • 5 years ago
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    there it went .....thought maye you fell asleep on your keyboard ;)

  3. anonymous
    • 5 years ago
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    sorry i am new took me a while to type the equation

  4. amistre64
    • 5 years ago
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    we can take out the constant "8" right?

  5. anonymous
    • 5 years ago
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    yes

  6. amistre64
    • 5 years ago
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    or do we want to use it later... if u = 4x^2 +1 ; then du = 8x dx dx = du/8x

  7. amistre64
    • 5 years ago
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    and x would equal... sqrt(u-1)/2...not sure if that helps us tho...

  8. anonymous
    • 5 years ago
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    well, this is supposed to be solved by using trig substitution. like,let 2x=tan\[\theta\]

  9. amistre64
    • 5 years ago
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    \[\int\limits{} \frac{2}{8u^2 \sqrt{u-1}} du\]

  10. amistre64
    • 5 years ago
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    we can try it by trig substitution....a little rusty, but why not :)

  11. amistre64
    • 5 years ago
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    4x^2 = tan^2(t) 2x = tan(t) ok

  12. amistre64
    • 5 years ago
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    tan^2 + 1 = sec^2 sec^2^2 = sec^4 right?

  13. amistre64
    • 5 years ago
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    1/sec^4 = cos^4 right?

  14. anonymous
    • 5 years ago
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    yes

  15. amistre64
    • 5 years ago
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    then when we pull out the 8 we get: [S] cos^4(t) dt to solve.... or did I miss somehting...

  16. amistre64
    • 5 years ago
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    it looks good to me :)

  17. anonymous
    • 5 years ago
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    wait, how did you get 4x^2 = tan^2t

  18. anonymous
    • 5 years ago
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    what i have after solving is cos^2t

  19. amistre64
    • 5 years ago
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    I simply defined it as such; since its a substitution, with the same value its good right?

  20. amistre64
    • 5 years ago
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    tan(t) = 2x -> t = tan^-1(2x)

  21. anonymous
    • 5 years ago
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    well, my teachet said a^2-x^2 : let x=asint a^2+x^2: x=atant and so on

  22. amistre64
    • 5 years ago
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    the value doesnt change, only the form it takes

  23. amistre64
    • 5 years ago
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    if we have something under a radical, then we want to substitute it with: asin(t) so that the "a" gets changed to a good value that can be taken away... there is no radical here to worry about so we dont have to worry about what "a" needs to be right?

  24. amistre64
    • 5 years ago
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    what we need here is not to get rid of a radical sign; but to transform it from one form to another.... we need tan^2 to equal 4x^2, the only way thats possible is to have tan = 2x

  25. amistre64
    • 5 years ago
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    makes sense?

  26. amistre64
    • 5 years ago
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    but I tend to miss something whenever I do this...that dt in the problem is not simply "dt" it is dt = something dx....and thats what I need to find lol

  27. amistre64
    • 5 years ago
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    tan(t) = 2x D (tan(t)) =D (2x) sec^2(t) dt = 2 dx dx = sec^2(t) dt/2 is what I should have... right?

  28. anonymous
    • 5 years ago
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    i will show u what i have. i just have problem at last

  29. amistre64
    • 5 years ago
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    so we end up with: 4 [S] cos^4 sec^2 dt

  30. amistre64
    • 5 years ago
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    ok :)

  31. amistre64
    • 5 years ago
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    4 [S] cos^2(t) dt is what it simplifies to.... im sure of it ;)

  32. anonymous
    • 5 years ago
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    yes

  33. amistre64
    • 5 years ago
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    cos^2 = 1 + sin^2..... and so on and so forth....

  34. amistre64
    • 5 years ago
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    we can reduce the power by uing the cos(2t) equations...

  35. anonymous
    • 5 years ago
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    yes what i cant do is do the final definite substitution

  36. amistre64
    • 5 years ago
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    cos(2t) = 2cos^2 - 1 1 + cos(2t) = 2cos^2 (1+cos(2t))/2 = cos^2 right?

  37. amistre64
    • 5 years ago
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    [S] 1/2 dt + [S] cos(2t)/2 dt .... looking more doable?

  38. amistre64
    • 5 years ago
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    \[4 \int\limits_{} \frac{1}{2} dt + \int\limits_{} \frac{\cos(2t)}{2}dt\]

  39. amistre64
    • 5 years ago
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    well that 4 times the whole lot of it :)

  40. amistre64
    • 5 years ago
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    the left side goes to 4t/2 = 2t what we need to do is concentrate in the right side...

  41. amistre64
    • 5 years ago
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    if we multiply it by 1 the value stays the same right?

  42. amistre64
    • 5 years ago
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    \[\int\limits_{} \frac{2}{2} * \frac{\cos(2t)}{2} dt\]

  43. amistre64
    • 5 years ago
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    lets use this to our advantage and take out the constants that dont matter.... ok?\[4* \frac{1}{4} \int\limits_{} 2 \cos(2t)dt\]

  44. amistre64
    • 5 years ago
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    this IS doable :)

  45. amistre64
    • 5 years ago
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    D (sin(2t)) = 2 cos(2t) right?

  46. anonymous
    • 5 years ago
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  47. amistre64
    • 5 years ago
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    very good :) make sure you keep track of the "4" we had and use it in both these integrals right?

  48. anonymous
    • 5 years ago
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    yea. the problem with me is the final part where we substiture value. cant figure that out

  49. amistre64
    • 5 years ago
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    so lets put up our solution as we have it so far...

  50. amistre64
    • 5 years ago
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    r\[F(x) = 2t + \sin(2t) + C\]

  51. amistre64
    • 5 years ago
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    you agree?

  52. anonymous
    • 5 years ago
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    yes

  53. amistre64
    • 5 years ago
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    this is our key then :)

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  54. anonymous
    • 5 years ago
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    wait, did we use pi/4 already?

  55. amistre64
    • 5 years ago
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    i didnt; I dont trust myself enough to change the original interval yet, so I just convert it back to values of x :)

  56. anonymous
    • 5 years ago
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    i am confused

  57. amistre64
    • 5 years ago
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    we can use this the way it is...minus the +C of course if we change the original interval to match oour substitution...which it looks like you might have down with that pi/4..... we can use that if your confident in it; but I prefer to undo the substitution of our trig with the key I provided...

  58. amistre64
    • 5 years ago
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    \[2 \tan^{-1}(2x) + \frac{4x}{4x^2+1}\] from [0, .5]

  59. amistre64
    • 5 years ago
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    if I did it right, my answer is 91 :)

  60. amistre64
    • 5 years ago
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    my tan^-1(1) came back as 45 instead of pi/4..... I might have to adjust for that :)

  61. amistre64
    • 5 years ago
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    (2+pi)/2 might be more appropriate....

  62. amistre64
    • 5 years ago
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    2.57 is about my answer .... any way to check it?

  63. amistre64
    • 5 years ago
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    2(pi/4) + sin(2pi/4) = pi/2 + 1 = (2+pi)/2 same results as before

  64. anonymous
    • 5 years ago
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    i will put that as answer. this was a test question . n way to check it. thanks for your time

  65. amistre64
    • 5 years ago
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    I hope it was helpful ;) thanx

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