calculus question is there a typo here,

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- anonymous

calculus question is there a typo here,

- schrodinger

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- anonymous

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- anonymous

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

- anonymous

No, it's not a typo.

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- anonymous

that doesnt make sense

- anonymous

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

- anonymous

Right. And dz =?

- anonymous

dz = 1/ cos z du ?

- anonymous

yes

- anonymous

why would they ask that , that seems strange

- anonymous

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

- anonymous

hmmm

- anonymous

thats redundant

- anonymous

Suit yourself. ;p

- anonymous

one sec, let me check

- anonymous

where exactly are you doing u substitution then

- anonymous

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/

- anonymous

you might as well just substitute it directly

- anonymous

so you get integral du / u^4

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I'm not sure what you mean by redundancy.

- anonymous

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

- anonymous

Sure, but sometimes it's not that obvious.

- anonymous

what do you mean

- anonymous

i NEVER use this approach

- anonymous

I always do.

- anonymous

give me problem and i will show yuo my approach

- anonymous

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.

- anonymous

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

- anonymous

everyone's brains work a little differently =)

- anonymous

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

- anonymous

give me a problem, lets see

- anonymous

integral x e^(x^2) dx

- anonymous

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

- anonymous

sure, fine.

- anonymous

ok :)

- anonymous

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

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