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Homework Help: Differentiation concept problem

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    i am suddenly stuck when differentiating. are my concepts right?

    if we integrate cosx/sinx, then since cosx is the differentiated form of sinx, then integrating d(sinx)/sinx = ln|sinx|

    so now, if i have : integrate z/(1-z2)1/2

    since z is the differentiated form of the denominator(barring the minus sign), (1-z2)1/2,
    then integrating d((1-z2)1/2)/(1-z2)1/2,

    shouldn't i get ln (1-z2)1/2 ? but the answer is without the ln, and true enough when i try to differentiate the ln (1-z2)1/2, i don't get back my original function.

    so where did i go wrong?
    thanks!
     
  2. jcsd
  3. Oct 15, 2011 #2

    SammyS

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    It's definitely true that [itex]\displaystyle \int\frac{d(\sqrt{1-z^2})}{\sqrt{1-z^2}}=\ln|\sqrt{1-z^2}|+C[/itex]

    But notice that [itex]\displaystyle d(\sqrt{1-z^2})=-\frac{z}{\sqrt{1-z^2}}dz\,.[/itex]

    So, [itex]\displaystyle \int\frac{d(\sqrt{1-z^2})}{\sqrt{1-z^2}}=-\int\frac{z\,dz}{\sqrt{1-z^2}}\,.[/itex]
     
  4. Oct 15, 2011 #3
    er i don't quite follow what you are saying?

    the answer given for [itex]\int\frac{z\,dz}{\sqrt{1-z^2}}\,.[/itex] is just - (1-z2)1/2

    if i understand what you are saying, you are implying that the answer should be ln(1-z2)1/2

    ???
     
  5. Oct 15, 2011 #4
    oh i see what you are trying to say.

    just that i think you have a typo at the last expression

    thanks!
     
  6. Oct 15, 2011 #5

    SammyS

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    Let's see:

    [itex]\displaystyle \frac{d}{dz}(\sqrt{1-z^2})=-\frac{z}{\sqrt{1-z^2}}\,.[/itex]

    So, the last one seems right.
     
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