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Help with indefinite integral

  1. Jul 5, 2010 #1
    1. The problem statement, all variables and given/known data
    The problem reads(from the 4th edition of Stewart Calculus "Concepts and Contexts" pg. 394 #17):

    Evaluate the integral.

    the indefinite integral of dx/x^2*sqrt(4-x^2)

    so this would read out as "the indefinite integral of dx over x squared times the square root of 4 minus x squared"


    2. Relevant equations




    3. The attempt at a solution


    I tried everything I know. I have been stuck on this problem for over 2 hours now and I just cant see what to do. The back of the book says the answer is

    -sqrt(4-x^2)/4x

    the frustrating thing is that using substitution I am getting to -sqrt(4-x^2)/x so I feel like I'm pretty close, or just have some careless mistake somewhere I am missing.
     
  2. jcsd
  3. Jul 5, 2010 #2

    rock.freak667

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    Try a trig substitution x=2sint
     
  4. Jul 5, 2010 #3
    I did already... but I guess I did it wrong.

    After the substitution I get to 1/4*integral sin^2t then do I go to the half-angle formula for sin^2t? By doing that I get 1/4*integral(1/2)(1-cos2t), then integrate to get 1/8*(t-.5sin2t).

    after that I solved for t using the trig substitution above to get t=arcsin(x/2). I then plug that into 1/8*(t-.5sin2t) to get 1/8*(arcsin(x/2)-.5sin2(arcsin(x/2), which I know is not right :/.
     
  5. Jul 5, 2010 #4

    Dick

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    To do the integral of 1/sin(t)^2 you are making it harder than it needs to be. You don't need any half angle formulas. What's the derivative of cot(t)? BTW write something like what you mean as 1/(4*sin^2(t)). Using parentheses will save you spelling it out in words.
     
    Last edited: Jul 5, 2010
  6. Jul 5, 2010 #5
    well I see that 1/sin(t)^2 =csc(t)^2, and the antiderivative of that is -cot(t)+c so i get all the way too

    -1/4*cot(t) where t=arcsin(x/2). Am I on the right track?
     
  7. Jul 5, 2010 #6

    Dick

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    If you can figure out why -1/4*cot(t) where t=arcsin(x/2) gives you -sqrt(4-x^2)/(4x) then you are all the way there. Yes, you are on the right track.
     
    Last edited: Jul 5, 2010
  8. Jul 5, 2010 #7
    alright thanks a lot. Its really late and I'd rather not spend 2 more hours stuck doing the wrong thing. Your(and others) assistance is much appreciated.
     
  9. Jul 5, 2010 #8

    Dick

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    You aren't doing the wrong thing! Probably wouldn't take 2 hours to figure out why cot(arcsin(x/2))=sqrt(1-x^2/4)/(x/2) but you are probably right. It will take you even less time in the morning.
     
  10. Jul 5, 2010 #9
    haha I know I'm not doing the wrong thing. I guess I meant to say im glad you told me I was on the right track because I wouldn't want to spend more time on this question tonight. 6 a.m. is calling my name for some calculus.
     
  11. Jul 5, 2010 #10

    Dick

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    Sleep agreeably knowing you'll get it the first thing tomorrow. You are SO CLOSE. Nite.
     
  12. Jul 6, 2010 #11
    Firstly it would be a whole lot easier to read if you wrote it in Latex like this:

    [tex]\int \frac{dx}{x^2 \cdot \sqrt{4-x^2}}[/tex]

    And I can give you the hint that in the chapter dealing with these types of integrals there is a hint on howto solve this integral!
     
    Last edited: Jul 6, 2010
  13. Jul 6, 2010 #12
    Cool! Where's the 'hint' located in the chapter?


    Yes I can see that Latex is much easier to read. I guess I'll work on learning to use Latex.
     
  14. Jul 6, 2010 #13

    hunt_mat

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    Here is an alternative:
    [tex]
    \int\frac{dx}{x^{2}\sqrt{4-x^{2}}}
    [/tex]
    Use [tex]x=2\cos u[/tex] to transform the integral to:
    [tex]
    -\frac{1}{4}\int\frac{du}{\cos^{2}u}=-\frac{1}{4}\int \sec^{2}udu
    [/tex]
    What well known function has derivative sec^{2}x?
    when you see a factor of [tex]\sqrt{a^{2}-x^{2}}[/tex] your first thought should be trig substitution.
     
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