Let [t] denote the greatest integer leq t. Th | vedclass

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(B) Let $I = \frac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\operatorname{cosec} x] - 5[\cot x]) \, dx$. First,consider $I_1 = \int_{\pi/6}^{5\pi/6} [\operato

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$14$

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(B) Let $I = \frac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\operatorname{cosec} x] - 5[\cot x]) \, dx$. First,consider $I_1 = \int_{\pi/6}^{5\pi/6} [\operatorname{cosec} x] \, dx$. In the interval $[\pi/6, 5\pi/6]$,$\operatorname{cosec} x \in [1, 2]$. Specifically,$[\operatorname{cosec} x] = 1$ for $x \in [\pi/6, 5\pi/6]$. Thus,$I_1 = \int_{\pi/6}^{5\pi/6} 1 \, dx = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$. Next,consider $I_2 = \int_{\pi/6}^{5\pi/6} [\cot x] \, dx$. Using the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$,we have $I_2 = \int_{\pi/6}^{5\pi/6} [\cot(\pi-x)] \, dx = \int_{\pi/6}^{5\pi/6} [-\cot x] \, dx$. Adding the two expressions for $I_2$: $2I_2 = \int_{\pi/6}^{5\pi/6} ([\cot x] + [-\cot x]) \, dx$. Using the property $[t] + [-t] = -1$ if $t 
otin \mathbb{Z}$ and $0$ if $t \in \mathbb{Z}$. Since $\cot x$ is an integer only at $x = \pi/2$,we have $[\cot x] + [-\cot x] = -1$ almost everywhere. $2I_2 = \int_{\pi/6}^{5\pi/6} (-1) \, dx = -(\frac{5\pi}{6} - \frac{\pi}{6}) = -\frac{2\pi}{3}$. So,$I_2 = -\frac{\pi}{3}$. Substituting back: $I = \frac{2}{\pi} (8 \cdot I_1 - 5 \cdot I_2) = \frac{2}{\pi} (8 \cdot \frac{2\pi}{3} - 5 \cdot (-\frac{\pi}{3})) = \frac{2}{\pi} (\frac{16\pi}{3} + \frac{5\pi}{3}) = \frac{2}{\pi} (\frac{21\pi}{3}) = \frac{2}{\pi} (7\pi) = 14$.

Let $[t]$ denote the greatest integer $\leq t$. Then $\frac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\operatorname{cosec} x] - 5[\cot x]) \, dx$ is equal to

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