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Confused by this integration

  1. Oct 15, 2015 #1
    So I was reading a solution of a problem that has to do with integration in polar coordinates, and in one of the steps it did this

    2yukcw1.png
    How did these terms (circled in blue) change like that? I've never seen a step like that before..
     
  2. jcsd
  3. Oct 15, 2015 #2

    Mark44

    Staff: Mentor

    What is the differential of ##\sin \theta##?

    If y = f(x), the differential of y is defined as dy = f'(x) dx
     
  4. Oct 15, 2015 #3

    pwsnafu

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    Science Advisor

    The substitution rule says
    ##\int f(x) \, \frac{du}{dx} \, dx = \int f(u(x)) \, du(x)##
    Hence, ##\theta = x##, ##\frac{du}{dx} = \cos\theta## and ##u = \sin \theta## gives
    ##\int f(\theta) \, cos(\theta) \, d\theta = \int f(\theta) \, d(\sin(\theta))##
     
  5. Oct 15, 2015 #4
    Why didn't they use the regular u-substitution, where u = cos(θ) , and du = -sin(θ) dθ ?
     
  6. Oct 15, 2015 #5

    mathman

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    Gold Member

    u = sin(θ) , and du = cos(θ) dθ avoids the complication of sign change.
     
  7. Oct 16, 2015 #6

    HallsofIvy

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    I'm not sure why you call that "the regular u-substitution". To integrate something like [itex]\int sin(\theta) cos(\theta) d\theta[/itex] either [itex]u= cos(\theta)[/itex], [itex]du= -sin(\theta)d\theta[/itex] or [itex]u= sin(\theta)[/itex], [itex]du= cos(\theta)d\theta[/itex] will work. And, I suspect that most people would use the latter since, as mathman said, it avoids the negative sign.
     
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