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Trig Substitution with Integration

  1. Dec 14, 2004 #1
    How would you go about solving
    [tex] \int \frac{\sqrt{1-x^2}}{x^2} [/tex] ?

    I have tried a few things... drawing out triangles... etc but can't seem to get it... I am kind of behind in math because I was gone for awhile because of being sick and presentations.
     
  2. jcsd
  3. Dec 14, 2004 #2

    Hurkyl

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    Ok, let's take a step back...

    Can you do [itex]\int \sqrt{1-x^2} \, dx[/itex]?
     
  4. Dec 14, 2004 #3

    dextercioby

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    Try the substitution [itex] x\rightarrow \sin u [/itex].And then part integration.
    Daniel
     
  5. Dec 14, 2004 #4
    Would that just be

    [tex]x=sin\theta[/tex]
    [tex] \sqrt{1-sin^2\theta [/tex]
    [tex] sin^2=1-cos^2\theta [/tex]
    [tex] \sqrt{1-(1-cos^2\theta) [/tex]
    [tex]\int \sqrt{-cos^2\theta [/tex]
     
    Last edited: Dec 14, 2004
  6. Dec 14, 2004 #5

    Hurkyl

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    Yep (assuming you meant cos2 θ, and get your signs right)
     
  7. Dec 14, 2004 #6
    hmm alright I am still lost even on your back up step
     
  8. Dec 14, 2004 #7

    Pyrrhus

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    Tom, again i emphatize on the important of the differential... do not forget about putting them on your integrals.

    Hurkyl means

    [tex] \sqrt{1 - \sin^2} = \sqrt{\cos^2}[/tex]
     
  9. Dec 14, 2004 #8
    As long as you're not in the middle of a test or something you can look it up in integral tables and see what answer they got and from that determine what methods they used. For example the solution to this integral is in a form that looks like it was done by parts, plus it has an inverse sine in it, which hints at trig substitutions as was mentioned by dextercioby.
     
  10. Dec 14, 2004 #9
    wow... i missed that competely i need more sleep... so what would I do to for the orignial problem with x^2 in the denominator
     
  11. Dec 14, 2004 #10

    Hurkyl

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    The same thing!
     
  12. Dec 14, 2004 #11
    Thats the problem I have a test on all the material I miseed comming up on thursday... and I need to make sure I get a good grade in the class if I want any chance of getting accepted after getting defred from MIT
     
  13. Dec 14, 2004 #12
    So would it just come out to be [tex] \int \frac{\sqrt{cos^2\theta}}{sin^2\theta}[/tex] ?
     
  14. Dec 14, 2004 #13

    dextercioby

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    Not exactly.U need to transform "dx" as well.That will give another "cosine".
    In should be
    [tex]\int \frac{\cos^{2}\theta}{\sin^{2}\theta}d\theta [/tex].
     
  15. Dec 14, 2004 #14
    oh yeah I forgot about that... so it becomes the [tex] \int tan^2\theta [/tex]
     
  16. Dec 14, 2004 #15
    My question is how did you decide to make [itex]x=sinu[/itex]
     
  17. Dec 14, 2004 #16
    That is a standard trig substitution, no?
     
  18. Dec 14, 2004 #17

    dextercioby

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    1.First of all it's [tex] \arctan^{2}\theta [/tex].
    2.Experience at doing integrals??? :wink: Actually it was the expression under the radical that led to the natural substitution "sine"/"cosine",just because:
    [tex] 1-\sin^{2}\theta =\cos^{2}\theta [/tex] and another one similar.

    Daniel.
     
    Last edited: Dec 14, 2004
  19. Dec 14, 2004 #18
    Alright I am going to assume [itex] x=4sinu [/itex] for

    [tex] \int \frac{x^3}{\sqrt{x^2-4}}dx [/tex]

    Therefore
    [tex]dx=cos\theta[/tex]

    [tex] \int \frac{4sin\theta^3}{\sqrt{4sin\theta^2-4}}*cos\theta [/tex]

    [tex]4 \int \frac{sin\theta^3}{\sqrt{sin^2\theta-1}} *cos\theta [/tex]

    [tex] 4 \int \frac{sin\theta^3}{\sqrt{cos^2}}*cos\theta [/tex]

    [tex] 4\int sin\theta^3 d\theta [/tex] ?????


    Did i do it right???
     
    Last edited: Dec 14, 2004
  20. Dec 14, 2004 #19
    it depends on what form it is in, if you have:
    (a^2 - x^2) : x = asin(theta)
    (a^2 + x^2) : x = atan(theta)
    (x^2 - a^2) : x = asec(theta)
     
  21. Dec 14, 2004 #20

    dextercioby

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    Hold on,what u posted is wrong:Û don't need 4,but 2 (check the denominator)
    [tex]dx=2 \cos\theta d\theta [/tex]
    [tex] x^{3}=8\sin^{3}\theta [/tex]

    Daniel.
     
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