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Intergration (1-(X^2))^0.5

  1. Oct 6, 2006 #1
    EDIT: my tex is a little broken trying to fix

    So i want to intergrate

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

    i start off by saying

    [tex]\sin u = x[/tex]

    so

    [tex]
    \frac {dx} {du} = \cos u
    [/tex]

    then

    [tex]
    \int (1 - x^2)^\frac{1}{2} \cos u du
    [/tex]

    which is

    [tex]
    \int \cos^2 u du
    [/tex]

    and

    [tex]
    \cos2u = 2cos^2 u - 1
    [/tex]

    so therefore

    [tex]
    \frac {1} {2} \cos2u + \frac {1} {2}= cos^2 u
    [/tex]

    so you intergrate

    [tex]
    \int \frac {1} {2} \cos 2u + \frac {1} {2} du
    [/tex]

    which is

    [tex]
    \frac {1} {4} \sin 2u + \frac {u} {2}
    [/tex]

    and

    [tex]
    \sin 2u = 2\sin u \cos u = 2\sin u(1 - sin^2 u)^\frac {1} {2}
    [/tex]

    putting x into u gets

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

    which im pretty sure is wrong. So can someone show me how to intergrate (1-(X^2))^0.5 ? i think that using x = sinu is wrong but u = sinx doesnt get me far either. Probably something simple ive overlooked.
     
    Last edited: Oct 6, 2006
  2. jcsd
  3. Oct 6, 2006 #2

    arildno

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    Absolute Values...
     
  4. Oct 6, 2006 #3
    thats correct
     
  5. Oct 6, 2006 #4
    Is that a subtle hint to stick the natural log function in there? because if you say

    [tex]
    u = 1 - x^2
    [/tex]

    then

    [tex]
    \int \frac{u^\frac{1}{2}}{(4 - 4u)^\frac{1}{2}} du
    [/tex]

    which looks sort of f'(x)/f(x) ish... ok tbh i have no clue why you said absolute value except it means always positive and is often in log when intergrating for some reason?
     
  6. Oct 6, 2006 #5

    arildno

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    [tex]\sqrt{x^{2}}=???[/tex]
     
  7. Oct 6, 2006 #6
    EDIT: tex is a work in progress... again :\

    I'm guessing due to the fact -x and x give the same answer

    [tex]\sqrt{x^{2}}=|x|[/tex]

    So

    [tex](1-(X^2))^\frac {1}{2} = (1-(|x| ^2))^\frac {1}{2} [/tex]

    and answer is

    [tex]
    \frac {x} {2} (1 - |x| ^2)^\frac {1} {2} + \frac {\cosec x} {2}
    [/tex]


    I'm worried about the fact trig has appeared in the intergral because f(x) didnt have any? And differential of cosec x is -cosecx cotx ???
     
    Last edited: Oct 6, 2006
  8. Oct 6, 2006 #7

    arildno

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    No..according to your (correct guess we must have:
    [tex](\cos^{2}(u))^{\frac{1}{2}}=|\cos(u)|[/tex]
    agreed?
     
  9. Oct 6, 2006 #8
    OK... so when i change cosu to sinu... since i have

    [tex](\cos^{2}(u))^{\frac{1}{2}}=(1 - \sin^{2}(u))^{\frac{1}{2}}[/tex]

    then i get

    [tex](|1 - sin{2}(u)|)^\frac{1}{2}[/tex] out of it?
    so the answer is

    [tex]
    \frac {x} {2} (|1 - x^{2}|)^\frac {1} {2} + \frac {\sin^{-1}(x)} {2}
    [/tex]
     
    Last edited: Oct 6, 2006
  10. Oct 6, 2006 #9

    arildno

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    I don't get what you're asking about; I have shown you that your flaw lies in your expression 5 in your original post.
     
  11. Oct 6, 2006 #10
    instead of [tex]
    \int \cos^2 u du
    [/tex]

    i should have

    [tex]
    \int \cos(u) | \cos(u) | du
    [/tex]

    ???
     
  12. Oct 6, 2006 #11

    arildno

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    That is correct!
     
  13. Oct 7, 2006 #12
    Ok so how do i intergrate

    [tex]
    \int \cos(u) | \cos(u) | du
    [/tex]

    ???

    btw i found a trial of a program called "Derive 6" on a demo disk i got from a maths course. It intergrated

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

    like this (click on thumbnail for bigger picture):

    [​IMG]

    Using that method can you avoid absolute values? I tried putting

    [tex]
    \int \cos(u) | \cos(u) | du
    [/tex]

    into it and it didnt like it much. Got stuck if you tried to get it to do it step by step. This is what it got if you went straight to the answer.

    [​IMG]




    I don't think it likes or uses absolute values much. When i told it to intergrate tanx it didnt use absolute values in the answer unlike my text book. So i'm hoping intergrating |cosx| isnt hard or complicated and the programs messing up?
     
  14. Oct 7, 2006 #13

    arildno

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    Well, the maximal domain of your x-integrand is -1 to 1.
    Divide your u-interval into those regions where the cosine is negative, and those where it is positive.
    You lose no generality by assumin [itex]0\leq{u}\leq{\pi}[/itex]
     
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