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ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2In other words, what is \(9\cos3t\) the derivative of?

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2of course with respect to \(t\).

Denebel
 2 years ago
Best ResponseYou've already chosen the best response.2Hint: \[\int\limits_{}^{} \cos ax dx = \frac{ \sin ax }{ a } + c\]

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2You start by taking the constant out.\[9\int\cos (3t)\]Use what Denebal gave.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2By the way, you can also take \(u = 3t\) :)

kylehberg
 2 years ago
Best ResponseYou've already chosen the best response.0would it be sin3/2t^2

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2\[9\int \cos(3t)dt\]Sorry for the mistake above

Chlorophyll
 2 years ago
Best ResponseYou've already chosen the best response.0Where do you get 2t^2 ???

kylehberg
 2 years ago
Best ResponseYou've already chosen the best response.0the antiderivate of 3t is 3/2t^2

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2\[9 \times{1 \over 3}\int \cos(u)du\]\[3\int \cos(u)du\]\[\implies 3\times \sin(u) + C\]

Denebel
 2 years ago
Best ResponseYou've already chosen the best response.2\[9\int\limits_{}^{} \cos (3t) dt = \frac{ 9\sin(3t) }{ 3 } = 3\sin(3t) + C\]

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.2Substituting \(u\) back,\[3\sin(3t) + C\]Yep

kylehberg
 2 years ago
Best ResponseYou've already chosen the best response.0thank you!... i can't do antiderivativs when they have a sin, cos, etc in them

Chlorophyll
 2 years ago
Best ResponseYou've already chosen the best response.0∫ cosu dx = sinu / u' Does it make any sense to you?

Chlorophyll
 2 years ago
Best ResponseYou've already chosen the best response.0or ∫cos ax dx = sin ax / a + C Does it make any sense at all?
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