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Math Help - Integral

  1. #1
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    Integral

    Attachment...

    Just wondering if this is correct..
    Attached Thumbnails Attached Thumbnails Integral-integral.jpg  
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  2. #2
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    no, it's actually \int_{0}^{\pi }{\sqrt{1-\cos ^{2}2x}\,dx}=\int_{0}^{\pi }{\left| \sin 2x \right|\,dx}.
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  3. #3
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    It's not correct, use 1-\cos^2 \theta = \sin^2\theta
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  4. #4
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    that was actually the first thing I tried however I got 0
    and the answer is suppose to be 2

    is there some kind of rule when using absolute values??
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  5. #5
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    \int_{0}^{\pi }{\left| \sin 2x \right|\,dx}=\frac{1}{2}\int_{0}^{2\pi }{\left| \sin x \right|\,dx}=\frac{1}{2}\left( \int_{0}^{\pi }{\sin x\,dx}-\int_{\pi }^{2\pi }{\sin x\,dx} \right)=2.
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  6. #6
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    o.O what?!? where did the 1/2 and 2pi come from and how did you remove the 2 from sin(2x)...???
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  7. #7
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    substitution t=2x.
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  8. #8
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    oh okay yeah it took me awhile cause you kept sinx as sinx

    but im still getting 0

    -cos(pi) = 1
    -cos(0) = -1

    --cos(2pi) = +1
    --cos(pi) = -1

    1-1+1-1 = 0
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