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Trig Substituion Integral

  1. Feb 26, 2008 #1
    Evaluate

    [tex]\int1/sqrt(4x^2-49)[/tex] for x>7/2

    Where I get lost really, is why do I set x = 7/2 sec u? The textbook just shows a generic formula where you always set x=a sec u. The only thing I could see is that anything less than 7/2 yiels a negative under the square root. But then again, this goes against the little formula which isn't really a problem, but take this other integral for example

    [tex]\int8dx/(4x^2+1)^2[/tex] here it shows setting x=1/2 tan u. But here I'm not really understanding the reason why. I'm guessing its because I'm not really sure why I set x equal to say tanget, sine, or whatever else.
     
    Last edited: Feb 26, 2008
  2. jcsd
  3. Feb 26, 2008 #2

    rock.freak667

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    Homework Helper

    If you take 2x=7sec[itex]\theta[/itex] then:

    [tex]2 dx=7 sec\theta tan\theta \theta[/tex]

    [tex]\int \frac{1}{\sqrt{4x^2-49}} dx [/tex]

    [tex]\equiv \int \frac{\frac{7}{2}sec \theta tan \theta}{\sqrt{49sec^2 \theta-49}[/tex]

    then use [itex] sec^2\theta-1=tan^2 \theta[/itex]

    for the 2nd integral. they use tan because when you substitute x=1/2 tanu then the denomination becomes

    (4(1/2 tanu)^2 +1)^2 = (tan^2u+1)^2 and well tan^u+1=sec^2 u
    so it becomes sec^4u.

    But basically what you want to do is make a substitution where the denominator (after substitution) can be used as another trig identity. e.g sin^2u+cos^2=1 etc.

    (EDIT: Not sure if my LaTex is showing up correctly[My browser is showing LaTex from questions I typed out many days ago!:confused:] so I don't know if you will understand what I wanted to say)
     
    Last edited: Feb 26, 2008
  4. Feb 26, 2008 #3
    Yes, it makes sense now. Its usually the little stuff that gets me all confused ^_^. If you take 2x=7sec is what I wasn't getting.
     
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