1. ## Tricky definite integral

Evaluate:

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

2. Originally Posted by jbuddenh
Evaluate:

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

3. 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 $({\pi},\frac{\pi}{2})$

$\arccos(cos(x))=\left\{\begin{array}{ccc}x & \mbox{for} & 0\leq x\leq \pi \\