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Integration using Euler Substitution

  1. Oct 30, 2014 #1
    Does anyone know of a derivation or justification of Euler's substitution formulas for evaluating irrational expressions? In other words, to evaluate integrals of the form:
    [tex]\int R(x,\sqrt{ax^2+bx+c})[/tex]
    You can use Euler's substitutions:
    [tex]1. \sqrt{ax^2+bx+c} = t \pm \sqrt{a}x, a>0 [/tex]
    [tex]2. \sqrt{ax^2+bx+c} = tx \pm \sqrt{c}, c>0 [/tex]
    [tex]3. \sqrt{ax^2+bx+c} = \sqrt{a(x-{x_{1}})(x-x_{2})}=t(x-x_{1})=t(x-x_{2}) [/tex]
    if [tex] x_{1}, x_{2} [/tex] are real.

    I understand how to mechanically evaluate the integrals but cannot find any justification or derivation of the substitution formulas, especially when to use, for example, t+x as opposed to t-x in the first instance.Can anyone help please? Thanks!
     
  2. jcsd
  3. Oct 30, 2014 #2
    They are actually not that different from declaring that x must equal k*sin(t) for some variable t when you want to make a trigonometric substitution. The intention of any substitution is to end up with a simpler algebraic expression, or at least an expression that you know can be integrated. We know that all rational functions can be integrated to a finite sum of polynomials, logarithms and arctangents. Euler's substitutions turn the square root of a quadratic into a rational function.
    Let's look at it this way. If the square root [itex]\sqrt{ax^2 + bx + c}[/itex] is a real number, then it is a number on the real line. It is therefore some real distance |t| away from [itex]\sqrt{a}[/itex], so we may write [itex]\sqrt{ax^2 + bx + c} = \sqrt{a} + t[/itex] for some number t. If we try to solve for the substitution function x(t) here, however, we get a horrific expression that does not simplify the integral (try it). However, if we try [itex]\sqrt{ax^2 + bx + c} = x\sqrt{a} + t[/itex], the problematic quadratic term in the squared equation disappears and we can solve for x easily, as a rational function of t. dx is then also a rational function, so we have solved the problem of finding a way to analytically compute the integral by reducing it to the integral of a rational function.
    The justification for most other rare substitutions would proceed along similar lines. Sometimes a substitution can be explained geometrically, which is pleasing, but many times it is just an attempt to replace x with any algebraic expression that simplifies the expression. The only requirement is that your substitution function be 1-1 on the interval of integration, and differentiable so that you can replace dx.
     
  4. Nov 3, 2014 #3
    So basically the Euler substitutions are tricks that work to simplify certain irrational expressions and don't have any formal derivations or geometric meanings?
     
  5. Nov 3, 2014 #4
    They certainly have formal derivations (the argument I gave can be made formal), and every algebraic expression can be given some geometric analogue. (For example, if we let [itex]y = \sqrt{ax^2 + bx + c}[/itex], then the first substitution amounts to [itex]y = x\sqrt{a} + t[/itex], which is the family of lines with slope [itex]\sqrt{a}[/itex] through points on the original curve, parametrized by the y-intercept t).
    However, they only apply to a very specific type of integral that is not generally encountered, so not many people have been interested in performing greater analysis on their form in as much depth as, say, elliptic integrals have received. I wouldn't really call them "tricks" as there is no surprise or misdirection involved. They are just a straightforward collection of related theorems.
     
    Last edited: Nov 3, 2014
  6. Mar 11, 2015 #5
    Draw a curve y2=ax2+bx+c
    Cut it with secant line
    If you choose proper points you will get second and third substitution
    If you cut the curve with line parallel to asymptote you will get first substitution
     
  7. Mar 11, 2015 #6
    If you are able to speak russian there is some explanation of geometrical meaning
    If you prefer polish language I can also send pdf in this language
     

    Attached Files:

    Last edited: Mar 11, 2015
  8. Mar 22, 2015 #7
    Unfortunately I don't speak Russian or Polish. Is there a recommended online site that would translate the document?
     
  9. Mar 27, 2015 #8
    if you pay for it you can find people able to translate it
    People are like this
    Automatized translations are not good quality
    I would translate it from polish to english but my english is not good
     
  10. Mar 27, 2015 #9
    I don't know of anyone that can speak good Russian or Polish. But I might have a better idea. I tried searching for some online translation services for PDF files and the document you attached is not editable or searchable or can be accessed by any free means. Can you please attach the first 2 pages as an RTF or DOC file with the formatting preserved? I bet that will work better.

    Thanks!
     
  11. Apr 2, 2015 #10
    Professional translation will cost you approximately 61 USD or 41 GBP
    I doubt that automatized translators are able to translate this because pdf was made from image file
    If you want to try rewrite it manually to doc file
     
  12. Apr 5, 2015 #11
    If you want to get u substitution for ths integrals draw a curve
    [tex]y^2=ax^2+bx+c[/tex]

    and cut it with secant line
    If secant line intersects curve at
    (λ,0)
    or
    (μ,0)
    you will get third substitution
    If secant line intersects curve at
    [tex](0,\sqrt{c})[/tex]
    or
    [tex](0,−\sqrt{c})[/tex]
    you will get second substitution

    Assume that
    a>0
    , draw asymptote and cut the curve with lines parallel to this asymptote
    This lines will intersect point
    (x0,y0)
    at infinity and point
    (x,y)
    which coordinates are rational functions of new variable
    You will get first substitution in this way

    This is my sketch of translation this document
    (I am not a professional , but more or less it is in this text)
     
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