If I = int_0^{frac{pi}{4}} log (1 + tan x) , | vedclass

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

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

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

If $I = \int_0^{\frac{\pi}{4}} \log (1 + \tan x) \, dx$,then the value of $I$ is

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