Question

FAN & MEDAL
calculus
question in comments

Answer 1

Integrate \[\int\limits_{}^{} \sin 3x dx\]

Answer 2

what function has derivative sine?

Answer 3

it just says that and to solve it using the u-substitution.

Answer 4

whatever it says, do you know a function whose derivative is sine?

Answer 5

no

Answer 6

Let u = 3x
du = 3dx
Substitute and integrate.

Answer 7

then there is no way on earth you can do this problem
but i really doubt you do not know it

Answer 8

think of a function whose derivative is sine
it is probably the first thing you think of

Answer 9

\[\frac{ 1 }{ 3 }du=dx\]?????

Answer 10

sure but that doesn't help you unless you know a function whose derivative is sine

Answer 11

hint: it is minus cosine

Answer 12

what if i substitute it into the original problem?

Answer 13

what you are looking for is a function whose derivative is \[\sin(3x)\] right?

Answer 14

so it has to be
\[-\cos(3x)\] but that is not quite right, because the derivative of
\[-\cos(3x)\] is
\[3\sin(3x)\] by the chain rule
you want
\[\sin(3x)\]so you have do divide by 3
in other words
\[\int \sin(3x)dx=-\frac{1}{3}\cos(3x)\]

Answer 15

the "u sub" you make (mentally) is \(u=3x, du=3dx.\frac{1}{3}du=dx\) and then get
\[\frac{1}{3}\int\sin(u)du\]

Answer 16

yea

Answer 17

\[=\frac{ 1 }{ 3 }(-\cos u)+C\]

Answer 18

\[=-\frac{ 1 }{ 3 }\cos 3x+C\]

Answer 19

would that be the answer?

Answer 20

@satellite73