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

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

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } I= \int_0^{\frac{\pi}{4}} \log (1+	an 	heta) \mathrm{d} 	heta \ &= \int_0^{\frac{\pi}{4}} \log \left[1+	an \left(\frac{\pi}{4}-	hetaight)ight] \mathrm{d} 	heta \ & \ldots\left[\because \int_0^a \mathrm{f}(x) \mathrm{d} x=\int_0^2 \mathrm{f}(\mathrm{a}-x) \mathrm{d} xight] \ &= \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-	an 	heta}{1+	an 	heta}ight) \mathrm{d} 	heta \ &=\int_0^{\frac{\pi}{4}} \log 2 \mathrm{~d} 	heta-\int_0^{\frac{\pi}{4}} \log (1+	an 	heta) \mathrm{d} 	heta \ & 	herefore \quad 2 \mathrm{I}=\int_0^{\frac{\pi}{4}} \log 2 \mathrm{~d} 	heta \Rightarrow \mathrm{I}=\frac{\log 2}{2}[	heta]_0^{\pi / 4}=\frac{\pi}{8} \log 2\end{aligned}$

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

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