If int_{0}^{pi} log (sin x) dx = 8 k,then int | vedclass

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(B) Let $I = \int_{0}^{\pi / 4} \log (1 + 	an x) dx$.Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:$I = \int_{0}^{\pi

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$-k$

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(B) Let $I = \int_{0}^{\pi / 4} \log (1 + 	an x) dx$.Using the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx$,we have:$I = \int_{0}^{\pi / 4} \log (1 + 	an (\frac{\pi}{4} - x)) dx$.Since $	an (\frac{\pi}{4} - x) = \frac{1 - 	an x}{1 + 	an x}$,we get:$I = \int_{0}^{\pi / 4} \log (1 + \frac{1 - 	an x}{1 + 	an x}) dx = \int_{0}^{\pi / 4} \log (\frac{2}{1 + 	an x}) dx$.$I = \int_{0}^{\pi / 4} (\log 2 - \log (1 + 	an x)) dx = \frac{\pi}{4} \log 2 - I$.$2I = \frac{\pi}{4} \log 2 \implies I = \frac{\pi}{8} \log 2$.Given $\int_{0}^{\pi} \log (\sin x) dx = 8k$. We know $\int_{0}^{\pi} \log (\sin x) dx = -\pi \log 2$.So,$8k = -\pi \log 2 \implies \pi \log 2 = -8k$.Substituting this into $I = \frac{\pi}{8} \log 2$,we get $I = \frac{1}{8} (-8k) = -k$.

If $\int_{0}^{\pi} \log (\sin x) dx = 8 k$,then $\int_{0}^{\pi / 4} \log (1 + \tan x) dx =$

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