int_{0}^{pi} [cot x] dx = | vedclass

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(C) Let $I = \int_{0}^{\pi} [\cot x] dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{\pi} [\cot(\pi - x

Vedclass · Accepted Answer

$-\frac{\pi}{2}$

Vedclass · Answer

(C) Let $I = \int_{0}^{\pi} [\cot x] dx$. Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we get: $I = \int_{0}^{\pi} [\cot(\pi - x)] dx = \int_{0}^{\pi} [-\cot x] dx$. Adding the two expressions for $I$: $2I = \int_{0}^{\pi} ([\cot x] + [-\cot x]) dx$. We know that for any $y 
otin \mathbb{Z}$,$[y] + [-y] = -1$,and for $y \in \mathbb{Z}$,$[y] + [-y] = 0$. The function $\cot x$ is not defined at $x=0, \pi/2, \pi$. At $x=\pi/2$,$\cot x = 0$,so $[0] + [0] = 0$. For all other $x \in (0, \pi)$,$\cot x$ is not an integer. Thus,$[\cot x] + [-\cot x] = -1$ almost everywhere in $(0, \pi)$. $2I = \int_{0}^{\pi} (-1) dx = -\pi$. Therefore,$I = -\frac{\pi}{2}$.

$\int_{0}^{\pi} [\cot x] dx = $

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