Question

calculus question! please help.
integrate: 2sin(0.03t)+1.5

Answer 1

is that (2*sin(.03))/(1.5)

?????

Answer 2

no.. it is 2*sin(0.03t) + 1.5

Answer 3

@satellite73 do you know how to do this? :)

Answer 4

yes

Answer 5

-2/.03cos(.03t)+1.5t+C

Answer 6

@sbc17 has it

Answer 7

@sbc17 could you please explain how you got that?

Answer 8

the anti derivative of \(1.5\) is \(1.5t\) that is clear right?

Answer 9

yes.

Answer 10

and the 2 is unimportant, it comes right out front of the integral

Answer 11

so the real question is, what is
\[\int\sin(.03t)dt\]

Answer 12

you know the antiderivative of \(\sin(t)\) is \(-\cos(t)\) so that part is okay yes?

Answer 13

yes.

Answer 14

therefore the real real question is
"how do you cope with the \(.03t\) inside the sine
and you do that via a mental u - sub
in your head you say \(u=.03t\) and so \(du=.03dt\) thus \(\frac{1}{.03}du = dt\) and you intergrate
\[\frac{1}{.03}\int\sin(u)du=-\frac{1}{3}\cos(u)=-\frac{1}{.03}\cos(.03t)\]

Answer 15

an easier example would be
\[\int\sin(5t)dt=-\frac{1}{5}\cos(5t)\]

Answer 16

thank you!