Question

$\int\limits_{0}^{\pi /4} \log (1 + \tan \, x ) dx$ is equal to :

• A. $\frac{\pi}{8} \log_{e} 2$
• B. $\frac{\pi}{4} \log_2 e$
• C. $\frac{\pi}{4} \log_e 2$
• D. $\frac{\pi}{8} \log_e \left( \frac{1}{2} \right)$

Answer 1

Answer: (A)

$\frac{\pi}{8} \log_{e} 2$

Answer 2

Let $I=\int\limits_{0}^{\pi / 4} \log (1+\tan x) d x...(1)$
$\Rightarrow I=\int\limits_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x$
$\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$=\int\limits_{0}^{\pi / 4} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x$
$=\int\limits_{0}^{\pi / 4} \log \left[\frac{2}{1+\tan x}\right] d x$
$=\int\limits_{0}^{\pi / 4} \log 2 dx -\int\limits_{0}^{\pi / 4} \log (1+\tan x ) d x$
$\Rightarrow I =\log 2[ x ]_{0}^{\pi / 4}- I [$ from E (i) $]$
$\Rightarrow 2 I =\frac{\pi}{4} \log _{ e } 2$
$\Rightarrow I =\frac{\pi}{8} \log _{ e } 2$