kylehberg
what is the antiderivate of 9cos3t???



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ParthKohli
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integrate it

ParthKohli
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In other words, what is \(9\cos3t\) the derivative of?

ParthKohli
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of course with respect to \(t\).

Denebel
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Hint: \[\int\limits_{}^{} \cos ax dx = \frac{ \sin ax }{ a } + c\]

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

ParthKohli
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Denebel*

ParthKohli
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By the way, you can also take \(u = 3t\) :)

kylehberg
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would it be sin3/2t^2

ParthKohli
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Erm, no.

ParthKohli
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\[9\int \cos(3t)dt\]Sorry for the mistake above

Chlorophyll
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Where do you get 2t^2 ???

kylehberg
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the antiderivate of 3t is 3/2t^2

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

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

ParthKohli
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Substituting \(u\) back,\[3\sin(3t) + C\]Yep

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

Chlorophyll
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∫ cosu dx = sinu / u'
Does it make any sense to you?

Chlorophyll
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or ∫cos ax dx = sin ax / a + C
Does it make any sense at all?

kylehberg
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yeah, thank you