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

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(D) 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^{\fra

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

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(D) 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(\frac{\pi}{4}-x) = \frac{	an(\frac{\pi}{4})-	an x}{1+	an(\frac{\pi}{4})	an x} = \frac{1-	an x}{1+	an x}$,we have: $I = \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-	an x}{1+	an x}ight) \, dx = \int_0^{\frac{\pi}{4}} \log \left(\frac{1+	an x+1-	an x}{1+	an x}ight) \, dx$. $I = \int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+	an x}ight) \, dx = \int_0^{\frac{\pi}{4}} (\log 2 - \log(1+	an x)) \, dx$. $I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - \int_0^{\frac{\pi}{4}} \log(1+	an x) \, dx$. $I = [x \log 2]_0^{\frac{\pi}{4}} - I$. $2I = \frac{\pi}{4} \log 2$. $I = \frac{\pi}{8} \log 2$.

$\int_0^{\frac{\pi}{4}} \log (1+\tan x) \, dx =$

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