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Question:

$\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$

Solution:

Let $I=\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$             ...(1)

$\therefore I=\int_{0}^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x$        $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left\{1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}\right\} d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \left\{1+\frac{1-\tan x}{1+\tan }\right\} d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log \frac{2}{(1+\tan x)} d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log 2 d x-\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \log 2 d x-I$        $[$ From (1) $]$

$\Rightarrow 2 I=[x \log 2]_{0}^{\frac{\pi}{4}}$

$\Rightarrow 2 I=\frac{\pi}{4} \log 2$

$\Rightarrow I=\frac{\pi}{8} \log 2$

 

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