Let $$[t]$$ denote the greatest integer $$\le t$$. Then $$\dfrac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\csc x] - 5[\cot x]) dx$$ is equal to ______.

We need to evaluate $$\dfrac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\csc x] - 5[\cot x])\, dx$$ where $$[t]$$ denotes the greatest integer $$\le t$$.

On the interval $$(\pi/6,5\pi/6)$$ the sine function is positive, so $$\csc x\ge1$$. At $$x=\pi/6$$ and $$x=5\pi/6$$ one finds $$\csc x=2$$, while at $$x=\pi/2$$ it attains the minimum value $$\csc x=1$$. Hence for almost all $$x\in(\pi/6,5\pi/6)$$ we have $$1\le\csc x\lt 2$$, and thus $$[\csc x]=1$$ throughout.

Meanwhile, $$\cot x$$ decreases continuously from $$\sqrt3$$ to $$-\sqrt3$$ as $$x$$ runs from $$\pi/6$$ to $$5\pi/6$$, crossing integer values at $$x=\pi/4,\pi/2,3\pi/4$$. On $$(\pi/6,\pi/4)$$ one has $$\cot x\in(1,\sqrt3)$$ so $$[\cot x]=1$$ and the length of this subinterval is $$\pi/4-\pi/6=\pi/12$$. On $$(\pi/4,\pi/2)$$ one has $$\cot x\in(0,1)$$ so $$[\cot x]=0$$ with length $$\pi/4$$. On $$(\pi/2,3\pi/4)$$ one has $$\cot x\in(-1,0)$$ so $$[\cot x]=-1$$ with length $$\pi/4$$. Finally on $$(3\pi/4,5\pi/6)$$ one has $$\cot x\in(-\sqrt3,-1)$$ so $$[\cot x]=-2$$ and the length is $$\pi/12$$.

Integrating $$[\csc x]$$ over $$[\pi/6,5\pi/6]$$ simply yields $$\int_{\pi/6}^{5\pi/6}[\csc x]\,dx=1\cdot\Bigl(\frac{5\pi}{6}-\frac{\pi}{6}\Bigr)=\frac{2\pi}{3}.$$

The integral of $$[\cot x]$$ splits into the four contributions: $$1\cdot\frac{\pi}{12}+0\cdot\frac{\pi}{4}+(-1)\cdot\frac{\pi}{4}+(-2)\cdot\frac{\pi}{12} =\frac{\pi}{12}-\frac{\pi}{4}-\frac{2\pi}{12} =-\frac{\pi}{3}.$$

Therefore $$I=\int_{\pi/6}^{5\pi/6}(8[\csc x]-5[\cot x])\,dx =8\cdot\frac{2\pi}{3}-5\cdot\bigl(-\tfrac{\pi}{3}\bigr) =\frac{16\pi}{3}+\frac{5\pi}{3} =7\pi,$$ and finally $$\frac{2}{\pi}\,I=\frac{2}{\pi}\cdot7\pi=14.$$