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krypton Group Title

y=d/dx integralcosx to 0 t/(1+t)dt

  • one year ago
  • one year ago

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  1. krypton Group Title
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    \[d/dx \int\limits_{0}^{cosx} \frac{ t }{ 1+t } dt\]

    • one year ago
  2. Outkast3r09 Group Title
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    \[\frac{cosx}{1+cosx}\]

    • one year ago
  3. krypton Group Title
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    how did u do it?

    • one year ago
  4. Zekarias Group Title
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    @Outkast3r09 I think u made a miss take in your first step (While u made the integral)

    • one year ago
  5. Outkast3r09 Group Title
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    the derivative of an integral is simply the integral itself with the variables in other words ... integrals and derivatives cancel eachother out

    • one year ago
  6. Outkast3r09 Group Title
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    \[\frac{cosx}{1+cosx} \]is your answer

    • one year ago
  7. Zekarias Group Title
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    It should be\[\cos(x)-\ln(1+\cos(x))\]

    • one year ago
  8. Outkast3r09 Group Title
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    |dw:1349804491261:dw|

    • one year ago
  9. Zekarias Group Title
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    Now take the derivative

    • one year ago
  10. Outkast3r09 Group Title
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    it shouldn't be anything @ zekarias

    • one year ago
  11. Outkast3r09 Group Title
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    @zekarias ... you don't need to do anything

    • one year ago
  12. Zekarias Group Title
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    What u get finally?

    • one year ago
  13. krypton Group Title
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    so the answer is just cosx/1+cosx ?

    • one year ago
  14. Outkast3r09 Group Title
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    yes

    • one year ago
  15. krypton Group Title
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    thanks,wolfram gives me the same answer,just wanted to make sure how it was done

    • one year ago
  16. Outkast3r09 Group Title
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    it's just simply knowing how an integral work... there is a lot of work to do it zekarias way but if you just know that the the derivative on an integral will bring you back to the f(x) within the integral... that's all you need to know(This works anytime the lower limit is some real number) because if you think about it.... |dw:1349804918350:dw| if you take the derivative of f(n)... you'll always get zero

    • one year ago
  17. krypton Group Title
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    oh thanks alot sweedy :)

    • one year ago
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