Question

How do I take the indefinite integral of sin(6x)?

Answer 1

how would you find \[\int sin(u)du\]?

Answer 2

or just \(\int sinx dx\)?

Answer 3

well the integral of \[\sin (x)\] is just \[-cos(x)\] , right?

Answer 4

yea so now, let's let u=6x what is the derivative of u?

Answer 5

\[u^2/2\]

Answer 6

the derivative of u*

Answer 7

sorry, du

Answer 8

so u=6x

(u=6x)'=?

Answer 9

*6dx

Answer 10

full math sentence speak(just humor me, I have a point, I promise)

Answer 11

u = 6x
du = 6xdx ?

Answer 12

perfect, now, could you solve that for dx?

Answer 13

du/6x = dx

Answer 14

ah, wait, typo from ya, du=6dx**

Answer 15

get rid of that x in the denominator and we are good

Answer 16

ah yes, sorry, it was a typo

Answer 17

So now, you have \[\int sin(6x)dx\]

Answer 18

du/6 = dx

Answer 19

I knew, from above, it's all good

Answer 20

so, if we say let u=6x and put that in, we need to sub out dx too, can you show me what that will look like?

Answer 21

I believe it would be \[\sin (u)du\]

Answer 22

well, what did dx= again?

Answer 23

you are close, just missing 1 important piece

Answer 24

\[\sin (u) * du/6\]

Answer 25

very good

Answer 26

now, I bet you can integrate that

Answer 27

\[-\cos6x/6?\]

Answer 28

yupp

Answer 29

Awesome, thanks.

Answer 30

np :)