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Theta ranges for trig substitution

  1. Sep 9, 2013 #1
    My professor, when doing trig substitution in lecture, always defines theta between certain intervals and when he takes the square root, he adds an absolute value bar to the trig function and then makes sure its positive through the interval. For practical purposes, is it necessary to go through all that hassle or might I trip up on a problem If I ignore theta ranges and take square roots normally, will I always get the right answer?
     
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  3. Sep 10, 2013 #2

    arildno

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    "If I ignore theta ranges and take square roots normally, will I always get the right answer? "

    No, you won't.

    Do as your professor does.
     
  4. Sep 10, 2013 #3
    can you give me an example? All the indefinite integrals I encountered, through his exercises as well as in high school, all worked without labeling the theta ranges.
     
  5. Sep 10, 2013 #4

    mathman

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    The indefinite integral of trig functions do not require absolute value. However, when there is a square root involved and you want to keep things real, you need absolute value.
     
  6. Sep 11, 2013 #5

    CAF123

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    There was a thread the other day where we wanted to find the length of the curve r=9+9cosθ.
    Following through the standard recipe of finding this, we arrive at the integral $$c \int_0^{2\pi} \sqrt{\cos^2\left(\frac{\theta}{2}\right)}\,d\theta,$$ where ##c## is a constant.

    Doing as you say, writing ##\sqrt{\cos^2\left(\frac{\theta}{2}\right)} = \cos\left(\frac{\theta}{2}\right)## will give a zero length, which is clearly false.
    The right way is of course ##\sqrt{\cos^2\left(\frac{\theta}{2}\right)} = |\cos\left(\frac{\theta}{2}\right)|## and this can be integrated by considering the sign of cos(θ/2) over the domain [0,2π].
     
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