int_{1}^{6pi}([sec^{-1}x]+[cot^{-1}x])dx is e | vedclass

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(D) We need to evaluate $I = \int_{1}^{6\pi} ([\sec^{-1}x] + [\cot^{-1}x]) dx$. Since $x \ge 1$,we have $0 < \sec^{-1}x < \pi/2$ and $0 < \cot^{-1}x <

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$6\pi-\sec 1$

Vedclass · Answer

(D) We need to evaluate $I = \int_{1}^{6\pi} ([\sec^{-1}x] + [\cot^{-1}x]) dx$. Since $x \ge 1$,we have $0 For $x \ge 1$,$\sec^{-1}x$ takes values in $[0, \pi/2)$. Thus,$[\sec^{-1}x] = 0$ for $1 \le x For $x \ge 1$,$\cot^{-1}x$ takes values in $(0, \pi/4]$. Thus,$[\cot^{-1}x] = 0$ for all $x \ge 1$. Therefore,the integral becomes $I = \int_{1}^{\sec 1} (0 + 0) dx + \int_{\sec 1}^{6\pi} (1 + 0) dx$. $I = 0 + [x]_{\sec 1}^{6\pi} = 6\pi - \sec 1$.

$\int_{1}^{6\pi}([\sec^{-1}x]+[\cot^{-1}x])dx$ is equal to (where $[.]$ denotes the greatest integer function).

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