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I Question on basic trig substitution with x = sin theta

  1. May 29, 2016 #1
    Say I have the integral of [ 1 / ( sqrt( 1 - x^2) ] * dx . Now I was told by many people in videos that I substitute x = sin theta, and this has me confused.

    Wouldn't I need to substitute x = cos theta instead? as x = cos theta on the unit circle instead of sin theta?

    Thanks in advance for your input! I am looking forward to hearing from you.
     
  2. jcsd
  3. May 29, 2016 #2

    Orodruin

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    It really does not matter. The substitution in itself has nothing to do with x being the x coordinate of the unit circle, it is just the name of a new integration variable. You can call it whatever you want, x, y, t, or ##\xi##, it is all the same.
     
  4. May 29, 2016 #3

    Math_QED

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    Orodruin's reply is a very good one. The only thing you use is the identity cos^2x + sin^2x = 1, so you can substitute both cos(t) or sin(t) for x.
     
  5. May 29, 2016 #4
    why is that even possible?

    I was watching an MIT Opencourseware lecture and the professor explicitly said that the trig substitutions are tied with x=cos theta and y = sin theta on the unit circle.

    The lecture is at:

    Thanks for your input.
     
  6. May 29, 2016 #5
    I understand if I substituted x = 1 - sin^2 x because that rests on the pythagorean identity, but that is obviously not the case.
     
  7. May 29, 2016 #6

    Orodruin

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    It is just a substitution of variables, you can use any invertible smooth function for this. The only thing is finding a function so that the resulting integral is simpler. This has nothing to do with polar coordinates or parametrising the unit circle - it is just a variable substitution
     
  8. May 29, 2016 #7

    micromass

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    Do you agree that

    $$\int \frac{dx}{\sqrt{1-x^2}} = \int \frac{dy}{\sqrt{1-y^2}}$$
     
  9. May 30, 2016 #8
    we just say in this case that x = y, so yes?
     
  10. May 30, 2016 #9
    but why do they float this idea round? it has to be based off the theory that x = rcos theta
     
  11. May 30, 2016 #10

    Orodruin

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    It has nothing to do with that. It is just a variable substitution.
     
  12. May 30, 2016 #11
    So how does this variable substitution work, then?
     
  13. May 30, 2016 #12

    Orodruin

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    The same way as any variable substitution. If you have an integral
    $$
    \int_a^b f(x) dx
    $$
    and introduce a one-to-one function ##x = g(t)##, then
    $$
    \int_a^b f(x) dx = \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(t)) g'(t) dt.
    $$
    In your case, ##f(x) = 1/\sqrt{1-x^2}## and ##g(t) = \sin(t)##.
     
  14. Jun 1, 2016 #13
    thank you for your help!
     
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