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Evaluating Integrals Using Trigonometric Function Substitutions Question

  1. Aug 8, 2011 #1
    Hi,

    I just had this idea pop into my head... Can you use a trig sub with a reference triangle who has sides equal to zero? or more like a value close to zero such as dx or da or something?

    For example

    integral 1/sqrt(9+dx^2) (dx)^2

    would have a reference triangle were the hypotenuse is sqrt(9+dx^2)
    one of the legs would be 3
    the other leg would be dx

    you then could express dx/sqrt(9+dx^2) as either csc(theta) or sec(theta) depending on what leg you set 3 be equal to...

    integral 1/sqrt(9+dx^2) (dx)^2 = integral csc(theta) dx
     
    Last edited: Aug 8, 2011
  2. jcsd
  3. Aug 9, 2011 #2
    I really wounder what can be done if one sets the side of the length of the triangle to dx and try and make a sub some how if that makes any sense
     
  4. Aug 9, 2011 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is an invalid integral. You cannot treat the "dx" as if it were a variable.
    An integral in x must be of the form [itex]\int f(x)dx[/itex].

     
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