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Math Help - Tricky definite integral

  1. #1
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    Tricky definite integral

    Evaluate:

    \int _{0}^{4\pi/3 }\!\arccos \left( \cos  x   \right) -<br />
x/2\,{dx}
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jbuddenh View Post
    Evaluate:

    \int _{0}^{4\pi/3 }\!\arccos \left( \cos  x   \right) -<br />
x/2\,{dx}
    obviously it is the \arccos ( \cos x ) part that is giving you trouble. so lets deal with the \int_0^{4 \pi/3} \arccos ( \cos x)~dx

    make a substitution, u = \cos x

    i leave it to you to show that this yields

    \int_{-1/2}^1 \frac {\arccos u}{\sqrt{1 - u^2}}~du

    and that integral shouldn't be too tricky for you. another substitution is required
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  3. #3
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    If you look at the graph of this, you see a triangle which is easy to find the area of.

    \frac{{\pi}\cdot\frac{\pi}{2}}{2}=\frac{{\pi}^{2}}  {4}

    and

    \frac{(\frac{4\pi}{3}-{\pi})\frac{\pi}{2})}{2}=\frac{{\pi}^{2}}{12}

    \frac{{\pi}^{2}}{4}+\frac{{\pi}^{2}}{12}=\frac{{\p  i}^{2}}{3}\approx 3.2898681337

    You can integrate the equivalent by just finding the area under the lines.

    \int_{0}^{\pi}\frac{x}{2}dx+\int_{\pi}^{\frac{4\pi  }{3}}(\frac{-3}{2}x+2{\pi})dx

    This will give you the same thing as that 'tough' integral.

    That is because the graph crosses the x-axis at \frac{4\pi}{3} and the vertex of the triangle is at ({\pi},\frac{\pi}{2})
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  4. #4
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    How about this representation then:

     \arccos(cos(x))=\left\{\begin{array}{ccc}x & \mbox{for} & 0\leq x\leq \pi \\<br />
                                     -x+2\pi &\mbox{for} &\pi\leq x\leq 2\pi<br />
                                     \end{array}\right.
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