Evulate : _0^ 4 (1+ x) d x. - Vidyadip

Question

Evulate :
$\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x.$

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Answer

$I=\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x \ldots(1)$
By property (6), $x=\frac{\pi}{4}-x$
$\begin{aligned}
I & =\int_0^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x \\
& =\int_0^{\frac{\pi}{4}} \log \left[1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \cdot \tan x}\right] d x \\
& =\int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x
\end{aligned}$
$\begin{aligned} I & =\int_0^{\frac{\pi}{4}} \log \left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right) d x \\ & =\int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) d x \\ & =\int_0^{\frac{\pi}{4}}(\log (2)-\log (1+\tan x)) d x\end{aligned}$
$I=\log 2 \int_0^{\frac{\pi}{4}} 1 d x-\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$
$\begin{array}{l} I =\log 2[x]_0^{\frac{\pi}{4}}- I \quad(\because \text { From equation }(1)) \\ 2 I =\log 2\left(\frac{\pi}{4}-0\right) \\ \therefore I =\frac{\pi}{8} \log 2\end{array}$

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