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inkyvoyd

  • 3 years ago

Integrate \(\frac{-\frac{1}{\sqrt{2}v}}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}\)

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  1. inkyvoyd
    • 3 years ago
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    @nbouscal will know what I am talking about, or maybe @Limitless

  2. inkyvoyd
    • 3 years ago
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    Wait, correction.

  3. inkyvoyd
    • 3 years ago
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    \(\frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}\)

  4. inkyvoyd
    • 3 years ago
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    Am I supposed to complete the square?

  5. inkyvoyd
    • 3 years ago
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    @dpaInc , pwease help me!

  6. dpaInc
    • 3 years ago
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    can't see it so i'm gonna copy it... and make it huge... \[\huge \frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1} \]

  7. inkyvoyd
    • 3 years ago
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    @dpaInc , how do you copy it like that?

  8. dpaInc
    • 3 years ago
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    right click-->show math as-->tex commands

  9. dpaInc
    • 3 years ago
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    \[\huge \int\frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}dv \]

  10. inkyvoyd
    • 3 years ago
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    Alright, now solve NAO!

  11. inkyvoyd
    • 3 years ago
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    xD

  12. inkyvoyd
    • 3 years ago
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    btw, this is the equation I got after trying 3 substitutions and seperating the fraction.

  13. dpaInc
    • 3 years ago
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    \[\large \int\frac{-\frac{1}{\sqrt{2}}v}{v^2+\sqrt{2}v+1}+\frac{\frac{1}{\sqrt{2}}v}{v^2-\sqrt{2}v+1}dv\] \[\large \frac{-1}{\sqrt2}\int\frac{v}{v^2+\sqrt{2}v+1}dv+\frac{1}{\sqrt2}\int \frac{v}{v^2-\sqrt{2}v+1}dv\] maybe i should just draw...

  14. inkyvoyd
    • 3 years ago
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    lolol

  15. inkyvoyd
    • 3 years ago
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    @dpaInc , now do I complete the square?

  16. dpaInc
    • 3 years ago
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    yeah... that's what i was about to do... but i think there is a formula... hang on....

  17. dpaInc
    • 3 years ago
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    there it is.... #13....:) http://integral-table.com/downloads/single-page-integral-table.pdf

  18. dpaInc
    • 3 years ago
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    i mean #16...l

  19. inkyvoyd
    • 3 years ago
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    \(\huge \frac{-1}{\sqrt2}\int\frac{v}{(v+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv+\frac{1}{\sqrt2}\int \frac{v}{(v-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv\)

  20. inkyvoyd
    • 3 years ago
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    But isn't using tables cheating?

  21. dpaInc
    • 3 years ago
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    no... of course not...

  22. inkyvoyd
    • 3 years ago
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    But... I don't know where the tables come from... >.<

  23. dpaInc
    • 3 years ago
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    it's like using all the different tires for your car... you're not going to make/invent your own tires are you? oh wait.. do you drive yet?

  24. inkyvoyd
    • 3 years ago
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    No lol. I'M GOING TO BUILD MY CAR FROM SCRATCH MUAHAHAHA

  25. dpaInc
    • 3 years ago
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    i don't think it's cheating... it's a good shortcut to cut through all that stuff..

  26. inkyvoyd
    • 3 years ago
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    Okay... @dpaInc , without showing me the math, can you tell me what to do after I have completed the square?

  27. dpaInc
    • 3 years ago
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    and as for me, i haven't really started using the integral tables until i saw everyon virtually used them here on OS.... i actually went through proving stuff because I don't use/memorize the tables...

  28. inkyvoyd
    • 3 years ago
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    @dpaInc , my problem is that I'm not sure how to get the results shown in the tables.

  29. dpaInc
    • 3 years ago
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    let me complete the square on the first integral... i haven't done it for a while and i guess i need the practice...

  30. inkyvoyd
    • 3 years ago
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    I already did in my last tex post :)

  31. inkyvoyd
    • 3 years ago
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    \huge \frac{-1}{\sqrt2}\int\frac{v}{(v+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv+\frac{1}{\sqrt2}\int \frac{v}{(v-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv

  32. dpaInc
    • 3 years ago
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    oh... i see you did it already....

  33. inkyvoyd
    • 3 years ago
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    xD

  34. nitz
    • 3 years ago
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    you are solving the problem in a wrong way

  35. inkyvoyd
    • 3 years ago
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    @nitz , what am I doing wrong?

  36. nitz
    • 3 years ago
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    see firstly make numerator as the derivative of denominator

  37. inkyvoyd
    • 3 years ago
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    Can you show me?

  38. nitz
    • 3 years ago
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    i am talking about 1st integration ie before addition

  39. inkyvoyd
    • 3 years ago
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    ?

  40. nitz
    • 3 years ago
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    in it, firstly multiply the numerator with 2 and add and subtract \[\sqrt{ 2} \] in it

  41. inkyvoyd
    • 3 years ago
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    You mean at the very start of the problem?

  42. nitz
    • 3 years ago
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    ya

  43. inkyvoyd
    • 3 years ago
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    but but but

  44. inkyvoyd
    • 3 years ago
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    \(\huge \frac{-1}{\sqrt2}\int\frac{v}{(v+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv+\frac{1}{\sqrt2}\int \frac{v}{(v-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dv\) we already got to here >.<

  45. nitz
    • 3 years ago
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    you cant solve it further in this case

  46. inkyvoyd
    • 3 years ago
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    #16? http://integral-table.com/downloads/single-page-integral-table.pdf

  47. nitz
    • 3 years ago
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    these are direct problems

  48. nitz
    • 3 years ago
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    these are direct solutions

  49. nitz
    • 3 years ago
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    you can apply direct formula

  50. dpaInc
    • 3 years ago
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    i think it works.... |dw:1340004704215:dw|

  51. inkyvoyd
    • 3 years ago
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    Uh, I can't use u in substiution LOL. I already used it...

  52. nbouscal
    • 3 years ago
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    Okay, I'm late to this party, but you can just complete the square and then do a substitution. This is just like the case that we reach in the integration of sqrt(tan x) that we have already done, inky.

  53. Limitless
    • 3 years ago
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    Next OS question: What made this integral appealing? I will medal the person with the best response.

  54. nbouscal
    • 3 years ago
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    This actually looks very similar to the integral of sqrt(tan x), if I had to guess I would say it is an intermediate step in a similar integration.

  55. inkyvoyd
    • 3 years ago
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    @nbouscal , that was my integration of the square root of tan x...

  56. Limitless
    • 3 years ago
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    nbouscal: Integral Detective

  57. inkyvoyd
    • 3 years ago
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    Maybe I should show what work i did before...

  58. nbouscal
    • 3 years ago
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    I didn't actually go back and look at that integral, you're probably right on. The next step is indeed to complete the square, then do another substitution.

  59. nbouscal
    • 3 years ago
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    Oh hey just look here: http://openstudy.com/study#/updates/4fb3c611e4b0556534298c6b

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