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Thread: Tricky definite integral

  1. #1
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    Tricky definite integral

    Evaluate:

    $\displaystyle \int _{0}^{4\pi/3 }\!\arccos \left( \cos x \right) -
    x/2\,{dx}$
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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:

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

    make a substitution, $\displaystyle u = \cos x$

    i leave it to you to show that this yields

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

    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.

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

    and

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

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

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

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

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

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

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