Question

calculus question is there a typo here,

Answer 1

number 2, where it says then dz = f(z) du,
shouldn't it say, then du = f(z) dz

Answer 2

No, it's not a typo.

Answer 3

that doesnt make sense

Answer 4

if u = sin z , right? then du = cos z dz ,

Answer 5

Right. And dz =?

Answer 6

dz = 1/ cos z du ?

Answer 7

yes

Answer 8

why would they ask that , that seems strange

Answer 9

Not really, that's how I solve all my u-subs.

Answer 10

hmmm

Answer 11

thats redundant

Answer 12

Suit yourself. ;p

Answer 13

one sec, let me check

Answer 14

where exactly are you doing u substitution then

Answer 15

u = sin z , then du = cos z dz , so if dz = du/ cos z we have
integral du/ u^4, oh the cos z cancels/

Answer 16

you might as well just substitute it directly

Answer 17

so you get integral du / u^4

Answer 18

this saves the step of having to plug that dz back in and then cancel, the whole point of u substitution i thought was to change the variables

Answer 19

\[u = sin\ z \implies du = (cos\ z)dz \implies dz = \frac{1}{cos\ z} du\]
\[\implies \int \frac{cos\ z}{sin^4z}dz = \int \frac{cos\ z}{u^4}(\frac{1}{cos\ z})du\]
\[ = \int \frac{1}{u^4} du \]

Answer 20

Sometimes it works nicely, othertimes you have to do a bit of finagling.

Answer 21

I'm not sure what you mean by redundancy.

Answer 22

its redundant, you can skip that step , ok because du = cos z dz, and you already have cos z dz in the numerator

Answer 23

Sure, but sometimes it's not that obvious.

Answer 24

what do you mean

Answer 25

i NEVER use this approach

Answer 26

I always do.

Answer 27

give me problem and i will show yuo my approach

Answer 28

I understand the other way, but too often it's just a matter of doing something in your head and I prefer to have it spelled out explicitly.

Answer 29

hmmm, it seems odd , i guess i learned it a different way

Answer 30

everyone's brains work a little differently =)

Answer 31

well let me show you what i do, and you compare it with what you do

Answer 32

give me a problem, lets see

Answer 33

integral x e^(x^2) dx

Answer 34

u = x^2 , du = 2x dx , du/2 = x dx (its ok to divide out constant)
so integral e^u du / 2

Answer 35

sure, fine.

Answer 36

ok :)

Answer 37

http://www.youtube.com/watch?v=Cj4y0EUlU-Y
About different ways people think.