Practicing Success
If $I=\int\limits_0^{\pi / 4} \log (1+\tan x) d x$, then $I=$ |
$\frac{\pi}{8} \log _e 2$ $\frac{\pi}{4} \log _e 2$ $-\frac{\pi}{8} \log _e 2$ $-\frac{\pi}{4} \log _e 2$ |
$\frac{\pi}{8} \log _e 2$ |
We have, $I=\int\limits_0^{\pi / 4} \log (1+\tan x) d x$ ....(i) $\Rightarrow I=\int\limits_0^{\pi / 4} \log \left\{1+\tan \left(\frac{\pi}{4}-x\right)\right\}$ [Using $\int\limits_0^a f(x)dx = \int\limits_0^a f(a-x)$] $\Rightarrow I=\int\limits_0^{\pi / 4} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x$ $\Rightarrow I=\int\limits_0^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x$ ....(ii) Adding (i) and (ii), we get $2 I=\int\limits_0^{\pi / 4} \log 2 d x=\frac{\pi}{4} \log _e 2 \Rightarrow I=\frac{\pi}{8} \log _e 2$ |