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By using the properties of definite integrals, evaluate the integrals
$\int_0^{\pi / 4} \log (1+\tan x) d x$
$\int_0^{\pi / 4} \log (1+\tan x) d x$
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Verified Answer
Let $\mathrm{I}=\int_0^{\pi / 4} \log (1+\tan \mathrm{x}) \mathrm{dx} \quad \ldots(i)$
Also $I=\int_0^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x$
$=\int_0^{\pi / 4} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x=\int_0^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x$
$\begin{aligned}
&=\int_0^{\pi / 4} \log 2 d x-\int_0^{\pi / 4} \log (1+\tan x) d x \\
&I=\log 2 \int_0^{\pi / 4} 1 d x-I
\end{aligned}$
$\Rightarrow 2 \mathrm{I}=\log 2[\mathrm{x}]_0^{\pi / 4}=\frac{\pi}{4} \log 2 \Rightarrow \mathrm{I}=\frac{\pi}{8} \log 2$
Also $I=\int_0^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x$
$=\int_0^{\pi / 4} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x=\int_0^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x$
$\begin{aligned}
&=\int_0^{\pi / 4} \log 2 d x-\int_0^{\pi / 4} \log (1+\tan x) d x \\
&I=\log 2 \int_0^{\pi / 4} 1 d x-I
\end{aligned}$
$\Rightarrow 2 \mathrm{I}=\log 2[\mathrm{x}]_0^{\pi / 4}=\frac{\pi}{4} \log 2 \Rightarrow \mathrm{I}=\frac{\pi}{8} \log 2$
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