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Math Help - How is it right? (definate integral)

  1. #1
    Senior Member polymerase's Avatar
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    How is it right? (definate integral)

    \int^{4\pi}_0 \sqrt{1+cos\;x}\;dx

    How is the answer 8\sqrt{2}? i keep getting 0!

    Thanks
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    \int^{4\pi}_0 \sqrt{1+cos\;x}\;dx

    How is the answer 8\sqrt{2}? i keep getting 0!

    Thanks
    Take a look at the graph. Since the integrand is never negative the area can't be 0. (The reason is that the square root of a number is never negative.)

    -Dan
    Attached Thumbnails Attached Thumbnails How is it right? (definate integral)-function.jpg  
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    Senior Member polymerase's Avatar
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    Quote Originally Posted by topsquark View Post
    Take a look at the graph. Since the integrand is never negative the area can't be 0. (The reason is that the square root of a number is never negative.)

    -Dan
    I understand it using a graph but i can't compute it algebraically thou....
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    I understand it using a graph but i can't compute it algebraically thou....
    Try
    cos \left ( \frac{1}{2}~x \right ) = \pm \sqrt{\frac{1 + cos(x)}{2}}

    The \pm is chosen by which quadrant your angle is in.

    -Dan
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    Quote Originally Posted by polymerase View Post
    \int^{4\pi}_0 \sqrt{1+cos\;x}\;dx

    How is the answer 8\sqrt{2}? i keep getting 0!

    Thanks
    1 + \cos x = 2\cos^2 \frac{x}{2}.

    This means, \int_0^{4\pi} \sqrt{1+\cos x} dx = \sqrt{2} \int_0^{4\pi}\left| \cos \frac{x}{2} \right| ~ dx

    Can you continue?
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