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Integrals of sqrt (simple, but I am getting frustrated!)

  1. Mar 20, 2005 #1
    I have been studying all day, and this little bugger keeps bothering me.
    How do I start this integral.
    int{sqrt(1-4x^2)}
    And what method do I use to solve it. Substitution or by Parts?

    :confused:
     
    Last edited: Mar 20, 2005
  2. jcsd
  3. Mar 20, 2005 #2
    can you think of some function [tex]f(u)[/tex] such that making the substitution [tex] x = f(u)[/tex] will make [tex]1-4x^2[/tex] into a perfect square?

    Think trig.
     
  4. Mar 20, 2005 #3
    Try using trig substitutions, if youve learned that already. Substitute x = (sinu)/4, and the integral will become int{sqrt(1-sinu^2)}dx,

    From there translate 1-sinu^2 to cosu^2 and go ahead and simplify sqrt(cos^2u).

    Dont forget since you made a change of variable you must change dx to du, do so by deriving your new x equation (x = sinu/4) and putting it into your integral. dx = (cosu/4)*du, du = 4dx/cosu

    So your integral will become int{cosu*4/cosu*du}. Your cosins will cancel and your final integral should become int{4*du}
     
    Last edited: Mar 20, 2005
  5. Mar 20, 2005 #4
    pft, sure, don't give her the chance to think about my question on her own!
     
  6. Mar 20, 2005 #5
    Sorry you weren't there when I posted my reply, you beat me to it :)
     
  7. Mar 20, 2005 #6
    And since you've posted that reply already, I should point out that you actually want to sub [tex] x = \frac{\sin{u}}{2}[/tex], not [tex] x = \frac{\sin{u}}{4}[/tex]! :smile:
     
  8. Mar 20, 2005 #7
    But how does that help me, because I still don't understand what process to use or where to start with the sqrt?
     
  9. Mar 20, 2005 #8
    By making the trig substitution, you will eliminate the square root because you will be square rooting cos^2(x). The square root of cos^2(x) is cos(x). After you are done with that part, you will have to convert the integral from dx to du by the steps above. After solving the integral you will have an expression in terms of u which you can solve for x by the relationship x = sinu/2
     
  10. Mar 20, 2005 #9
    It's a method called "trigonometric substitution."

    Let's see what happens when we sub in [itex] x = \frac{\sin{u}}{2} \Longrightarrow dx = \frac{\cos{u}}{2}du[/itex]:

    we get

    [tex] \sqrt{1 - 4x^2} = \sqrt{1 - 4\frac{\sin^2{u}}{4}} = \sqrt{1 - \sin^2{u}}[/tex]

    Recalling that [itex] \cos^2{u} = 1 - \sin^2{u}[/itex], this just turns into

    [tex]\sqrt{\cos^2{u}} = \cos{u}[/tex]

    Thus your integral is

    [tex]\int \sqrt{1 - 4x^2} dx = \int \cos{u} dx[/tex]

    and substituting in for [itex]dx[/itex], this is just

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

    and I'm sure you can work it out from there :)

    As I mentioned, this is an example of a method called trigonometric substitution. Some other examples are integrands involving [itex]\sqrt{a^2 + x^2}[/itex] in which we substitute [itex] x = a\tan{u}[/itex] to give [itex]\sqrt{a^2 + a^2\tan^2{u}} = \sqrt{a^2\sec^2{u}} = a\sec{u}[/itex], or involving [itex] \sqrt{x^2 - a^2}[/itex] where we sub [itex] x = a\sec{u}[/itex] to make a similar simplification.
     
  11. Mar 20, 2005 #10
    ahh, you beat me this time!
     
  12. Mar 21, 2005 #11

    dextercioby

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    Instead of using sec & tan,u can very nicely use [itex] \sinh,\cosh [/itex]...

    Hyperbolic trigonometry is useful & elegant.
     
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