int_{0}^{pi} log(sin^2 x) , dx = | vedclass

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(A) Let $I = \int_{0}^{\pi} \log(\sin^2 x) \, dx$. Using the property $\log(a^b) = b \log a$,we have $I = \int_{0}^{\pi} 2 \log(\sin x) \, dx$. Since

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$2\pi \log_e\left(\frac{1}{2}\right)$

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(A) Let $I = \int_{0}^{\pi} \log(\sin^2 x) \, dx$. Using the property $\log(a^b) = b \log a$,we have $I = \int_{0}^{\pi} 2 \log(\sin x) \, dx$. Since $\sin x$ is symmetric about $x = \frac{\pi}{2}$ in the interval $[0, \pi]$,we can write $I = 2 	imes 2 \int_{0}^{\pi/2} \log(\sin x) \, dx = 4 \int_{0}^{\pi/2} \log(\sin x) \, dx$. Using the standard integral result $\int_{0}^{\pi/2} \log(\sin x) \, dx = -\frac{\pi}{2} \log 2$,we get: $I = 4 	imes \left(-\frac{\pi}{2} \log 2ight) = -2\pi \log 2$. Since $-\log 2 = \log(2^{-1}) = \log(\frac{1}{2})$,we have $I = 2\pi \log(\frac{1}{2})$.

$\int_{0}^{\pi} \log(\sin^2 x) \, dx = $

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