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krypton Group Title

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

  • 2 years ago
  • 2 years ago

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

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

    • 2 years ago
  3. krypton Group Title
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    how did u do it?

    • 2 years 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)

    • 2 years 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

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

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

    • 2 years ago
  8. Outkast3r09 Group Title
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    |dw:1349804491261:dw|

    • 2 years ago
  9. Zekarias Group Title
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    Now take the derivative

    • 2 years ago
  10. Outkast3r09 Group Title
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    it shouldn't be anything @ zekarias

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

    • 2 years ago
  12. Zekarias Group Title
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    What u get finally?

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

    • 2 years ago
  14. Outkast3r09 Group Title
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    yes

    • 2 years 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

    • 2 years 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

    • 2 years ago
  17. krypton Group Title
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    oh thanks alot sweedy :)

    • 2 years ago
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