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

ParthKohliBest ResponseYou've already chosen the best response.2
of course with respect to \(t\).
 one year ago

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

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

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

kylehbergBest ResponseYou've already chosen the best response.0
would it be sin3/2t^2
 one year ago

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

ChlorophyllBest ResponseYou've already chosen the best response.0
Where do you get 2t^2 ???
 one year ago

kylehbergBest ResponseYou've already chosen the best response.0
the antiderivate of 3t is 3/2t^2
 one year ago

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

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

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

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

ChlorophyllBest ResponseYou've already chosen the best response.0
∫ cosu dx = sinu / u' Does it make any sense to you?
 one year ago

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