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kylehberg

  • 3 years ago

what is the antiderivate of 9cos3t???

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  1. ParthKohli
    • 3 years ago
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    integrate it

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

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

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

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

  6. ParthKohli
    • 3 years ago
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    Denebel*

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

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

  9. ParthKohli
    • 3 years ago
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    Erm, no.

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

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

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

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

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

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

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

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

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

  19. kylehberg
    • 3 years ago
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    yeah, thank you

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