Question
By using the properties of definite integrals, evaluate the integral $\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$

Answer

Given $\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$
Let $I=\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$ .....(i)
as, $\left\{\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right\}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-\mathrm{x}\right)\right] \mathrm{d} \mathrm{x}$
as $\left\{\tan (A-B)=\frac{\tan (A)-\tan (B)}{1+\tan (A) \tan (B)}\right\}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left[1+\frac{\tan \left(\frac{\pi}{4}\right)-\tan (\mathrm{x})}{1+\tan \left(\frac{\pi}{4}\right) \tan (\mathrm{x})}\right] \mathrm{dx}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log \left[1+\frac{1-\tan (\mathrm{x})}{1+\tan (\mathrm{x})}\right] \mathrm{d} \mathrm{x}$
$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left[\frac{2}{1+\tan (x)}\right] d x$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log [2] \mathrm{d} \mathrm{x}-\int_{0}^{\frac{\pi}{4}} \log [1+\tan (\mathrm{x})] \mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log [2] \mathrm{d} \mathrm{x}-\mathrm{I}$ (from (i))
$\Rightarrow 2 \mathrm{I}=[\mathrm{x} \log 2]_{0}^{\frac{\pi}{4}}$
$\Rightarrow 2 \mathrm{I}=\frac{\pi}{4} \log 2-0$
$\Rightarrow \mathrm{I}=\frac{\pi}{8} \log 2$

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