int_0^{pi / 4} log (1+tan x) d x= | vedclass

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(C) Let $I = \int_0^{\pi / 4} \log (1+	an x) d x$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = \int_0^{\pi / 4} \log \left

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$\frac{\pi}{8} \log 2$

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(C) Let $I = \int_0^{\pi / 4} \log (1+	an x) d x$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = \int_0^{\pi / 4} \log \left[1+	an \left(\frac{\pi}{4}-xight)ight] d x$.Since $	an(\frac{\pi}{4}-x) = \frac{1-	an x}{1+	an x}$,the integral becomes:$I = \int_0^{\pi / 4} \log \left[1+\frac{1-	an x}{1+	an x}ight] d x = \int_0^{\pi / 4} \log \left(\frac{1+	an x + 1 - 	an x}{1+	an x}ight) d x$.$I = \int_0^{\pi / 4} \log \left(\frac{2}{1+	an x}ight) d x$.Using the property $\log(\frac{a}{b}) = \log a - \log b$,we get:$I = \int_0^{\pi / 4} \log 2 d x - \int_0^{\pi / 4} \log (1+	an x) d x$.$I = \int_0^{\pi / 4} \log 2 d x - I$.$2I = \log 2 \int_0^{\pi / 4} d x = \log 2 [x]_0^{\pi / 4} = \frac{\pi}{4} \log 2$.Therefore,$I = \frac{\pi}{8} \log 2$.

$\int_0^{\pi / 4} \log (1+\tan x) d x=$

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