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Mathematics 34 Online
OpenStudy (anonymous):

calculus question is there a typo here,

OpenStudy (anonymous):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

No, it's not a typo.

OpenStudy (anonymous):

that doesnt make sense

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Right. And dz =?

OpenStudy (anonymous):

dz = 1/ cos z du ?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

why would they ask that , that seems strange

OpenStudy (anonymous):

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

OpenStudy (anonymous):

hmmm

OpenStudy (anonymous):

thats redundant

OpenStudy (anonymous):

Suit yourself. ;p

OpenStudy (anonymous):

one sec, let me check

OpenStudy (anonymous):

where exactly are you doing u substitution then

OpenStudy (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/

OpenStudy (anonymous):

you might as well just substitute it directly

OpenStudy (anonymous):

so you get integral du / u^4

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I'm not sure what you mean by redundancy.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Sure, but sometimes it's not that obvious.

OpenStudy (anonymous):

what do you mean

OpenStudy (anonymous):

i NEVER use this approach

OpenStudy (anonymous):

I always do.

OpenStudy (anonymous):

give me problem and i will show yuo my approach

OpenStudy (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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

everyone's brains work a little differently =)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

give me a problem, lets see

OpenStudy (anonymous):

integral x e^(x^2) dx

OpenStudy (anonymous):

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

OpenStudy (anonymous):

sure, fine.

OpenStudy (anonymous):

ok :)

OpenStudy (anonymous):

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

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