The value of int_{0}^{pi / 2} log (operatorna | vedclass

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Question

(A) Let $I = \int_{0}^{\pi / 2} \log (\operatorname{cosec} x) d x$. Since $\operatorname{cosec} x = \frac{1}{\sin x}$,we have $\log (\operatorname{cos

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$\frac{\pi}{2} \log 2$

Vedclass · Answer

(A) Let $I = \int_{0}^{\pi / 2} \log (\operatorname{cosec} x) d x$. Since $\operatorname{cosec} x = \frac{1}{\sin x}$,we have $\log (\operatorname{cosec} x) = \log (\sin x)^{-1} = -\log \sin x$. Therefore,$I = -\int_{0}^{\pi / 2} \log \sin x d x$. Using the standard definite integral result $\int_{0}^{\pi / 2} \log \sin x d x = -\frac{\pi}{2} \log 2$,we get: $I = -(-\frac{\pi}{2} \log 2) = \frac{\pi}{2} \log 2$.

The value of $\int_{0}^{\pi / 2} \log (\operatorname{cosec} x) d x$ is

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