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Math Help - split integral of abs + trig.

  1. #1
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    split integral of abs + trig.

    Hi,

    I need to express the following integral without the abs.

    \int_0^\pi |\frac{1}{2} + Cos(x)|dx.

    I know that Cos(x) = \frac{1}{2} when x = \frac{\pi}{3} which I am sure is part of the solution.

    How do I express this without the absolute value.
    What if it wasn't a nice fraction like \frac{1}{2} what if it was

    \int_0^\pi |\frac{3}{7} + Cos(x)|dx.

    How would you solve it then.


    Thanks
    Regards
    Craig.
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  2. #2
    Moo
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    Hello,

    See on a unit circle, how cos fluctuates between 0 and \pi : it's always decreasing. (in another way, the arccos function is strictly decreasing in this interval)

    So if you're studying \int_0^\pi |a+\cos(x)| ~dx, you have to see where a+\cos(x)<0, that is \cos(x)<{\color{red}-}a

    Thanks to this unit circle, it's easy (and by knowing that arccos is strictly decreasing, it's easy too) : \forall x\in[0,\pi] ~,~ \cos(x)<-a \Leftrightarrow x>\arccos(-a)

    So we'll have \int_0^\pi |a+\cos(x)| ~dx=\int_0^{\arccos(-a)} a+\cos(x) ~dx-\int_{\arccos(-a)}^\pi a+\cos(x) ~dx
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