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Math Help - Stuck on this trig integral

  1. #1
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    Stuck on this trig integral

    integrate square root of 1-sin2xdx , [o, pie/4]
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  2. #2
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    Hello, twilightstr!

    \int^{\frac{\pi}{4}}_0 \sqrt{1-\sin2x}\,dx
    Under the radical, multiply by: \frac{1+\sin2x}{1+\sin2x}

    . . \sqrt{\frac{1-\sin2x}{1}\cdot\frac{1+\sin2x}{1+\sin2x}} \;=\;\sqrt{\frac{1-\sin^2\!2x}{1+\sin2x}} \;=\;\sqrt{\frac{\cos^2\!2x}{1 + \sin2x}}\;=\;\frac{\cos2x}{\sqrt{1+\sin2x}}


    And we have: . \int^{\frac{\pi}{4}}_0 (1 + \sin2x)^{-\frac{1}{2}}(\cos2x\,dx)

    Then let: u \:=\:1 + \sin2x


    Got it?

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  3. #3
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    \sqrt{{{\cos }^{2}}2x}=\left| \cos 2x \right|=\cos 2x, because this last is positive for \left[ 0,\frac{\pi }{4} \right].
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  4. #4
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    Another way is to recognize that:

    1 - \sin{(2x)} = \sin^2{x} - 2\sin{x}\cos{x} + \cos^2{x} = (\sin{x} - \cos{x})^2

    Also take note that:
    \sqrt{(\sin{x} - \cos{x})^2} = |\sin{x} - \cos{x}| = -(\sin{x}-\cos{x}) for \left[ 0, \frac{\pi}{4} \right]
    Last edited by Krizalid; February 7th 2009 at 07:07 AM.
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