1. We need to evaluate $I=\frac{2}{\pi}\int_{\pi/6}^{5\pi/6}\left(8[\csc x]-5[\cot x]\right)dx.$ Here $[t]$ denotes the greatest integer function.

2. First analyze $[\csc x]$ on $\left[\frac{\pi}{6},\frac{5\pi}{6}\right]$.

Since $\sin x\in \left[\frac12,1\right]$ on this interval, we get $\csc x=\frac1{\sin x}\in [1,2].$ Now:

• At $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$, $\sin x=\frac12$, so $\csc x=2$ and hence $[\csc x]=2$.
• For all interior points, $1<\csc x<2$, so $[\csc x]=1$.

The endpoints do not affect the integral, so effectively $[\csc x]=1\quad \text{almost everywhere on } \left[\frac{\pi}{6},\frac{5\pi}{6}\right].$ Thus $\int_{\pi/6}^{5\pi/6}8[\csc x]dx=8\left(\frac{5\pi}{6}-\frac{\pi}{6}\right)=8\cdot \frac{2\pi}{3}=\frac{16\pi}{3}.$

3. Now analyze $[\cot x]$ on the interval.

Since $\cot x$ decreases continuously from $\cot\frac{\pi}{6}=\sqrt3$ to $\cot\frac{5\pi}{6}=-\sqrt3,$ we locate where it crosses integers:

• $\cot x=1 \Rightarrow x=\frac{\pi}{4}$
• $\cot x=0 \Rightarrow x=\frac{\pi}{2}$
• $\cot x=-1 \Rightarrow x=\frac{3\pi}{4}$

Hence:

• For $x\in\left[\frac{\pi}{6},\frac{\pi}{4}\right)$, $1\le \cot x\le \sqrt3$, so $[\cot x]=1$.
• For $x\in\left[\frac{\pi}{4},\frac{\pi}{2}\right)$, $0\le \cot x<1$, so $[\cot x]=0$.
• For $x\in\left[\frac{\pi}{2},\frac{3\pi}{4}\right)$, $-1\le \cot x<0$, so $[\cot x]=-1$.
• For $x\in\left[\frac{3\pi}{4},\frac{5\pi}{6}\right]$, $-\sqrt3\le \cot x<-1$, so $[\cot x]=-2$ except at the single point $x=\frac{3\pi}{4}$ where it equals $-1$; this single point does not affect the integral.

Therefore,