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anonymous

  • 5 years ago

calculus question is there a typo here,

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  1. anonymous
    • 5 years ago
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  2. anonymous
    • 5 years ago
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    number 2, where it says then dz = f(z) du, shouldn't it say, then du = f(z) dz

  3. anonymous
    • 5 years ago
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    No, it's not a typo.

  4. anonymous
    • 5 years ago
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    that doesnt make sense

  5. anonymous
    • 5 years ago
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    if u = sin z , right? then du = cos z dz ,

  6. anonymous
    • 5 years ago
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    Right. And dz =?

  7. anonymous
    • 5 years ago
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    dz = 1/ cos z du ?

  8. anonymous
    • 5 years ago
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    yes

  9. anonymous
    • 5 years ago
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    why would they ask that , that seems strange

  10. anonymous
    • 5 years ago
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    Not really, that's how I solve all my u-subs.

  11. anonymous
    • 5 years ago
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    hmmm

  12. anonymous
    • 5 years ago
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    thats redundant

  13. anonymous
    • 5 years ago
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    Suit yourself. ;p

  14. anonymous
    • 5 years ago
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    one sec, let me check

  15. anonymous
    • 5 years ago
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    where exactly are you doing u substitution then

  16. anonymous
    • 5 years ago
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    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/

  17. anonymous
    • 5 years ago
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    you might as well just substitute it directly

  18. anonymous
    • 5 years ago
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    so you get integral du / u^4

  19. anonymous
    • 5 years ago
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    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

  20. anonymous
    • 5 years ago
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    \[u = sin\ z \implies du = (cos\ z)dz \implies dz = \frac{1}{cos\ z} du\] \[\implies \int \frac{cos\ z}{sin^4z}dz = \int \frac{cos\ z}{u^4}(\frac{1}{cos\ z})du\] \[ = \int \frac{1}{u^4} du \]

  21. anonymous
    • 5 years ago
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    Sometimes it works nicely, othertimes you have to do a bit of finagling.

  22. anonymous
    • 5 years ago
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    I'm not sure what you mean by redundancy.

  23. anonymous
    • 5 years ago
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    its redundant, you can skip that step , ok because du = cos z dz, and you already have cos z dz in the numerator

  24. anonymous
    • 5 years ago
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    Sure, but sometimes it's not that obvious.

  25. anonymous
    • 5 years ago
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    what do you mean

  26. anonymous
    • 5 years ago
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    i NEVER use this approach

  27. anonymous
    • 5 years ago
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    I always do.

  28. anonymous
    • 5 years ago
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    give me problem and i will show yuo my approach

  29. anonymous
    • 5 years ago
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    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.

  30. anonymous
    • 5 years ago
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    hmmm, it seems odd , i guess i learned it a different way

  31. anonymous
    • 5 years ago
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    everyone's brains work a little differently =)

  32. anonymous
    • 5 years ago
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    well let me show you what i do, and you compare it with what you do

  33. anonymous
    • 5 years ago
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    give me a problem, lets see

  34. anonymous
    • 5 years ago
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    integral x e^(x^2) dx

  35. anonymous
    • 5 years ago
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    u = x^2 , du = 2x dx , du/2 = x dx (its ok to divide out constant) so integral e^u du / 2

  36. anonymous
    • 5 years ago
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    sure, fine.

  37. anonymous
    • 5 years ago
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    ok :)

  38. anonymous
    • 5 years ago
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    http://www.youtube.com/watch?v=Cj4y0EUlU-Y About different ways people think.

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