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Math Help - Simple way to do an integral

  1. #1
    Member billa's Avatar
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    Simple way to do an integral

    I am trying to do ...

    \int_0^{2\pi}\sqrt{1-cos(t)}dt

    Here is what I did. Can you either comment on my work or post a better solution?

    Assume the domain of all functions is (0,\pi). Let u=1-cos(t)

    \displaystyle\int\sqrt{1-cos(x)}dx=\int\frac{sin(x)}{\sqrt{1+cos(x)}}dx=\in  t\frac{1}{\sqrt{2-u}}du=-2\sqrt{1+cos(x)}

    Therefore, \int_0^\pi\sqrt{1-cos(x)}dx=2\sqrt{2}

    Since the integrand is symmetric about y=\pi,

    \int_0^{2\pi}\sqrt{1-cos(t)}dt=2*2\sqrt{2}
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Santiago, Chile
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    well the answer is right, but your procedure seems to be a little bit messy.

    we use the fact that \cos t=1-2{{\sin }^{2}}\dfrac{t}{2}, hence your integral becomes \displaystyle2\sqrt{2}\int_{0}^{\pi }{\left| \sin t \right|\,dt}=2\sqrt{2}\int_{0}^{\pi }{\sin t\,dt}=4\sqrt{2}.
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  3. #3
    Member billa's Avatar
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    Awesome! Many thanks, Krizalid.
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