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jaersyn

  • 3 years ago

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)

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  1. Dido525
    • 3 years ago
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    Oh my god... This is a monster to integrate...

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

  3. jaersyn
    • 3 years ago
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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

  4. Dido525
    • 3 years ago
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    You don't have to use intergration by parts. You can use substitution twice.

  5. jaersyn
    • 3 years ago
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    ?

  6. Dido525
    • 3 years ago
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    Nevermind. I am wrong.

  7. Dido525
    • 3 years ago
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    Dammit. Sorry. I can't solve this :( . It's beyond my level of math.

  8. jaersyn
    • 3 years ago
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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

  9. Dido525
    • 3 years ago
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    I don't think so... Sorry. I can't help you any further with this :( .

  10. jaersyn
    • 3 years ago
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    that's okay, thanks for looking though

  11. jaersyn
    • 3 years ago
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    i know igbasallote got dis

  12. jaersyn
    • 3 years ago
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    level 99 math wizard

  13. Dido525
    • 3 years ago
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    I would love to see how to do this question.

  14. RadEn
    • 3 years ago
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    I GOT IT, BUT VERY LONG TO WRITTING THIS STEPS :(

  15. RadEn
    • 3 years ago
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    as information (if im not mistake), i got the answer : |dw:1349498667471:dw|

  16. RadEn
    • 3 years ago
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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

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