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Indefinite Quotients

  1. Jul 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the indefinite integral of

    [tex]\int \frac{sec^2 x}{\sqrt{1-tan^2 x}} dx [/tex]


    I'm stuck on how to proceed with the denominator. I know that

    [tex]\sqrt{1+tan^2 x}[/tex]

    is equivalent to [tex]\\sec\theta[/tex]

    but I cant seen to find an equivalent to what I have. Can anyone give me any pointers?
     
    Last edited: Jul 12, 2007
  2. jcsd
  3. Jul 12, 2007 #2

    Dick

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    What's the derivative of arcsin?
     
  4. Jul 12, 2007 #3

    George Jones

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    Can you make a sustitution involving [itex]u[/itex] such that [itex]du = sec^2 x dx[/itex]?
     
  5. Jul 12, 2007 #4
    The derivative of arcsin would be

    [tex]\frac{1}{(1 - x^2)^\frac{1}{2}}[/tex]
     
  6. Jul 12, 2007 #5

    Dick

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    Right. Suppose we put tan(x) in place of x?
     
  7. Jul 12, 2007 #6
    This would give us

    [tex]\frac{1}{(1 - tan^2 (x))^\frac{1}{2}}[/tex]

    Which leaves us with a term equivalent to us getting rid of the square root
     
  8. Jul 12, 2007 #7

    Dick

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    d/dx(arcsin(tan(x)))=?. Don't forget the chain rule.
     
    Last edited: Jul 12, 2007
  9. Jul 12, 2007 #8
    Hmm... This is where I'm losing it I think. The chain rule will give us

    [tex]f(x)=arcsin[/tex]
    [tex]f'(x)=\frac{1}{(1-x^2)^\frac{1}{2}}[/tex]
    [tex]g(x)=tan(x)[/tex]
    [tex]g'(x)=1+tan^2 (x)[/tex]


    [tex]\frac{d}{dx}arcsin(tan(x))=\frac{1}{(1-x^2)^\frac{1}{2}} \times tan(x) \times \left( 1+tan^2 (x) \right)[/tex]

    Not sure if this is right but I've provided my values so you can see where I've gone wrong
     
    Last edited: Jul 12, 2007
  10. Jul 12, 2007 #9

    Dick

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    I'm sorry. I think I'm going way too fast and confusing you. George's step by step suggestion is probably much better. The chain rule is (f(g(x)))'=f'(g(x))*g'(x). You have written f'(x)*g(x)*g'(x). Do you see the difference? And there is a more compact way to write 1+tan(x)^2. When you finish this go back and look at the route George would have led you on.
     
    Last edited: Jul 12, 2007
  11. Jul 12, 2007 #10
    Ok, is there a quick way of identifying what needs to be substituted. At the moment I'm doing it by trial and error, and obviously in an exam, thats not going to be very economical
     
  12. Jul 12, 2007 #11

    Dick

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    You want something whose derivative is sec^2(x). Memorizing at least small list of derivative and integral forms is necessary to do this kind of stuff with any speed. You've already said that the derivative of tan(x) is tan^2(x)+1. This is certainly correct, but most people would probably respond with a different (but equivalent) answer. What does the wikipedia table of derivatives say?
     
  13. Jul 12, 2007 #12
    Just had a look at wikipedia. It shows that the more common answer for the derivative of tan(x) is

    [tex]sec^2 (x)[/tex]

    or

    [tex]\frac{1}{cos^2 (x)}[/tex]
     
  14. Jul 12, 2007 #13

    Dick

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    Exactly, so if u=tan(x) then du=sec^2(x)*dx.
     
  15. Jul 12, 2007 #14
    Ok, I'm happy with how we got to this point.

    So what I have now is the integral of 'something' times sec^2(x) dx

    Is this right? And now I need to find what makes up the equivalent of the original integral.... Am I on the right lines here?

    I think half the problem I'm having with this question is that my books don't give such detailed examples. I need to follow a lot more examples... Do you know of any good sites (apart from this one) that gives plenty of worked examples?
     
    Last edited: Jul 12, 2007
  16. Jul 12, 2007 #15

    Dick

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    The point is that u=tan(x) might be a good substitution to do on the original integral.
     
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