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kylehberg Group Title

what is the antiderivate of 9cos3t???

  • 2 years ago
  • 2 years ago

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  1. ParthKohli Group Title
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    integrate it

    • 2 years ago
  2. ParthKohli Group Title
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    In other words, what is \(9\cos3t\) the derivative of?

    • 2 years ago
  3. ParthKohli Group Title
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    of course with respect to \(t\).

    • 2 years ago
  4. Denebel Group Title
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    Hint: \[\int\limits_{}^{} \cos ax dx = \frac{ \sin ax }{ a } + c\]

    • 2 years ago
  5. ParthKohli Group Title
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    You start by taking the constant out.\[9\int\cos (3t)\]Use what Denebal gave.

    • 2 years ago
  6. ParthKohli Group Title
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    Denebel*

    • 2 years ago
  7. ParthKohli Group Title
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    By the way, you can also take \(u = 3t\) :)

    • 2 years ago
  8. kylehberg Group Title
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    would it be sin3/2t^2

    • 2 years ago
  9. ParthKohli Group Title
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    Erm, no.

    • 2 years ago
  10. ParthKohli Group Title
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    \[9\int \cos(3t)dt\]Sorry for the mistake above

    • 2 years ago
  11. Chlorophyll Group Title
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    Where do you get 2t^2 ???

    • 2 years ago
  12. kylehberg Group Title
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    the antiderivate of 3t is 3/2t^2

    • 2 years ago
  13. ParthKohli Group Title
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    \[9 \times{1 \over 3}\int \cos(u)du\]\[3\int \cos(u)du\]\[\implies 3\times \sin(u) + C\]

    • 2 years ago
  14. Denebel Group Title
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    \[9\int\limits_{}^{} \cos (3t) dt = \frac{ 9\sin(3t) }{ 3 } = 3\sin(3t) + C\]

    • 2 years ago
  15. ParthKohli Group Title
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    Substituting \(u\) back,\[3\sin(3t) + C\]Yep

    • 2 years ago
  16. kylehberg Group Title
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    thank you!... i can't do antiderivativs when they have a sin, cos, etc in them

    • 2 years ago
  17. Chlorophyll Group Title
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    ∫ cosu dx = sinu / u' Does it make any sense to you?

    • 2 years ago
  18. Chlorophyll Group Title
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    or ∫cos ax dx = sin ax / a + C Does it make any sense at all?

    • 2 years ago
  19. kylehberg Group Title
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    yeah, thank you

    • 2 years ago
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