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what is the antiderivate of 9cos3t???

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integrate it
In other words, what is \(9\cos3t\) the derivative of?
of course with respect to \(t\).

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Hint: \[\int\limits_{}^{} \cos ax dx = \frac{ \sin ax }{ a } + c\]
You start by taking the constant out.\[9\int\cos (3t)\]Use what Denebal gave.
By the way, you can also take \(u = 3t\) :)
would it be sin3/2t^2
Erm, no.
\[9\int \cos(3t)dt\]Sorry for the mistake above
Where do you get 2t^2 ???
the antiderivate of 3t is 3/2t^2
\[9 \times{1 \over 3}\int \cos(u)du\]\[3\int \cos(u)du\]\[\implies 3\times \sin(u) + C\]
\[9\int\limits_{}^{} \cos (3t) dt = \frac{ 9\sin(3t) }{ 3 } = 3\sin(3t) + C\]
Substituting \(u\) back,\[3\sin(3t) + C\]Yep
thank you!... i can't do antiderivativs when they have a sin, cos, etc in them
∫ cosu dx = sinu / u' Does it make any sense to you?
or ∫cos ax dx = sin ax / a + C Does it make any sense at all?
yeah, thank you

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