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jaersyn Group Title

integrate 1/(5+sinx-3cosx) let u = tan(x/2) cosx = (1-u^2)/(1+u^2) sinx = 2u/(1+u^2) dx = 2/(1+u^2)

  • one year ago
  • one year ago

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  1. Dido525 Group Title
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    Oh my god... This is a monster to integrate...

    • one year ago
  2. jaersyn Group Title
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    yeah, after substituting everything i got up to 1/[4(u+(1/8))^2 + (15/64)]

    • one year ago
  3. jaersyn Group Title
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    if you could help me with that, that'd be cool. I don't know what to do next to integrate

    • one year ago
  4. Dido525 Group Title
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    You don't have to use intergration by parts. You can use substitution twice.

    • one year ago
  5. jaersyn Group Title
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    ?

    • one year ago
  6. Dido525 Group Title
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    Nevermind. I am wrong.

    • one year ago
  7. Dido525 Group Title
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    Dammit. Sorry. I can't solve this :( . It's beyond my level of math.

    • one year ago
  8. jaersyn Group Title
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    quick unrelated question: if i have 1/u(1-u^2) can i use partial fractions to integrate? even though its (-u^2 + 1) in the denom? i never came across a negative like that

    • one year ago
  9. Dido525 Group Title
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    I don't think so... Sorry. I can't help you any further with this :( .

    • one year ago
  10. jaersyn Group Title
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    that's okay, thanks for looking though

    • one year ago
  11. jaersyn Group Title
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    i know igbasallote got dis

    • one year ago
  12. jaersyn Group Title
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    level 99 math wizard

    • one year ago
  13. Dido525 Group Title
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    I would love to see how to do this question.

    • one year ago
  14. RadEn Group Title
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    I GOT IT, BUT VERY LONG TO WRITTING THIS STEPS :(

    • one year ago
  15. RadEn Group Title
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    as information (if im not mistake), i got the answer : |dw:1349498667471:dw|

    • one year ago
  16. RadEn Group Title
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    yes, i think my answer is right ^^,, bacasue the "god of math" told me that : http://www.wolframalpha.com/input/?i=int%28dx%2F%285%2Bsin%28x%29-3cos%28x%29%29%29

    • one year ago
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