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
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 $\displaystyle ({\pi},\frac{\pi}{2})$