Question

what is the antiderivate of 9cos3t???

Answer 1

integrate it

Answer 2

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

Answer 3

of course with respect to \(t\).

Answer 4

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

Answer 5

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

Answer 6

Denebel*

Answer 7

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

Answer 8

would it be sin3/2t^2

Answer 9

Erm, no.

Answer 10

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

Answer 11

Where do you get 2t^2 ???

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

∫ cosu dx = sinu / u'

Does it make any sense to you?

Answer 18

or ∫cos ax dx = sin ax / a + C

Does it make any sense at all?

Answer 19

yeah, thank you