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Math Help - Integrating sqrt(1-sin2x)

  1. #1
    Junior Member gusztav's Avatar
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    Integrating sqrt(1-sin2x)

    Hello,

    I'd really appreciate some help with finding the following indefinite integral:

    \int \sqrt{1- \sin 2x} dx


    What I tried is the following:

    \int \sqrt{1- \sin 2x} dx=

    \int \sqrt{1- 2\sin x \cos x} dx=

    \int \sqrt{\sin^2x + \cos^2 x - 2\sin x \cos x} dx=

    \int \sqrt{\sin^2x - 2\sin x \cos x + \cos^2 x } dx=

    \int \sqrt{(\sin x -  \cos x )^2} dx=


    but it would be WRONG to proceed like this:
    \int \sqrt{(\sin x -  \cos x )^2} dx=
    \int (\sin x -  \cos x ) dx=
    \int \sin x dx -  \int \cos x dx=
    -\cos x -\sin x +C


    because \sqrt{a^2} could be either a or -a !


    So perhaps this problem needs some other approach.



    Mathematica gives this:

    \frac {(\cos x + \sin x)\sqrt {1 - \sin  2 x}} {\cos x - \sin x}

    as the answer.

    Thanks a lot for your help!
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  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
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    Jordan
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    1,093
    you answer is correct

    proof

    (1-sin2x)^1/2=sinx-cosx your solution substitute it in



    \frac {(\cos x + \sin x)\sqrt {1 - \sin  2 x}} {\cos x - \sin x}<br />


    you get your answer
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