anonymous
  • anonymous
calculus question is there a typo here,
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
number 2, where it says then dz = f(z) du, shouldn't it say, then du = f(z) dz
anonymous
  • anonymous
No, it's not a typo.

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anonymous
  • anonymous
that doesnt make sense
anonymous
  • anonymous
if u = sin z , right? then du = cos z dz ,
anonymous
  • anonymous
Right. And dz =?
anonymous
  • anonymous
dz = 1/ cos z du ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
why would they ask that , that seems strange
anonymous
  • anonymous
Not really, that's how I solve all my u-subs.
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
thats redundant
anonymous
  • anonymous
Suit yourself. ;p
anonymous
  • anonymous
one sec, let me check
anonymous
  • anonymous
where exactly are you doing u substitution then
anonymous
  • 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
  • anonymous
you might as well just substitute it directly
anonymous
  • anonymous
so you get integral du / u^4
anonymous
  • 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
  • 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
  • anonymous
Sometimes it works nicely, othertimes you have to do a bit of finagling.
anonymous
  • anonymous
I'm not sure what you mean by redundancy.
anonymous
  • 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
  • anonymous
Sure, but sometimes it's not that obvious.
anonymous
  • anonymous
what do you mean
anonymous
  • anonymous
i NEVER use this approach
anonymous
  • anonymous
I always do.
anonymous
  • anonymous
give me problem and i will show yuo my approach
anonymous
  • 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
  • anonymous
hmmm, it seems odd , i guess i learned it a different way
anonymous
  • anonymous
everyone's brains work a little differently =)
anonymous
  • anonymous
well let me show you what i do, and you compare it with what you do
anonymous
  • anonymous
give me a problem, lets see
anonymous
  • anonymous
integral x e^(x^2) dx
anonymous
  • anonymous
u = x^2 , du = 2x dx , du/2 = x dx (its ok to divide out constant) so integral e^u du / 2
anonymous
  • anonymous
sure, fine.
anonymous
  • anonymous
ok :)
anonymous
  • anonymous
http://www.youtube.com/watch?v=Cj4y0EUlU-Y About different ways people think.

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