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

Question

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

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Answer

$\text { Let } I =\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x\ldots(i)$
$=\int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x \quad \ldots . . .\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$=\int_0^{\frac{\pi}{4}} \log \left(1+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \cdot \tan x}\right) d x$
$=\int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x$
$=\int_0^{\frac{\pi}{4}} \log \left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right) d x$
$=\int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) d x$
$=\int_0^{\frac{\pi}{4}}[\log 2-\log (1+\tan x)] d x$
$=\log 2 \int_0^{\frac{\pi}{4}} 1 \cdot d x-\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$
$\left.\therefore I =\log 2[x]_0^{\frac{\pi}{4}}- I \quad \text {......[From (i) }\right]$
$\therefore 2 l =\log 2\left(\frac{\pi}{4}-0\right)$
$\therefore I =\frac{\pi}{8} \log 2$

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