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Math Help - How do I evaluate the integral of (sin(2t)^2)^(1/2) ?

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    How do I evaluate the integral of (sin(2t)^2)^(1/2) ?

    I had to evaluate an integral from 0 to 2pi in which the integrand was sqrt(sin(2x)^2). I figured the square root of sin(2x) squared was just sin(2x). Evaluating the integral gave me 0, which makes sense since the function is symmetric. But this was "incorrect".

    Apparently sqrt(sin(2x)^2) = |sin(2x)|. How would I express the integral of the absolute value of a function? (Please generalize your answer so that it explains cases for which the function is not symmetric.)

    |sin(2x)| = sin(2x) and -sin(2x) so would the integral of |sin(2x)| from 0 to 2pi be the sum of the integral of sin(2x) and the integral of -sin(2x) from 0 to 2pi?
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    Re: How do I evaluate the integral of (sin(2t)^2)^(1/2) ?

    It should be obvious that the sum of sin(2x) and -sin(2x) is 0 and so the sum of the integrals will be 0!

    The basic definition of "absolute value" is that |x|= x if x\ge 0, -x is x< 0.

    So |sin(2x)|= sin(2x) if 0\le 2x\le \pi and -sin(2x) if \pi\le 2x\le 2\pi which is the same as
    |sin(2x)|= sin(2x) if [tex]0\le x\le \pi/2[/itex] and -sin(2x) if \pi/2\le x\le \pi

    If you want to integrate from 0 to 2\pi, you will also need to use the fact that, sine is periodic with period 2\pi, sin(x) is positive for 2\pi\<  x< 3\pi and negative for 3\pi< x< 4\pi so |sin(2x)|= sin(2x) for \pi< x< 3\pi/2 and |sin(2x)|= -sin(2x) for 3\pi/2< x< 2\pi
    Last edited by HallsofIvy; April 26th 2012 at 12:31 PM.
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