Question

$\int\limits_0^{\pi/4}\log (\frac {\sin \,x+\cos\,x}{\cos\,x})dx $ is equal to

• A. $\frac {\pi} {4}\log\, 2$
• B. $\frac {\pi} {2}\log\, 2$
• C. $\frac {\pi} {8}\log\, 2$
• D. $\log\,2$

Answer 1

Answer: (C)

$I =\int\limits_{0}^{\pi / 4} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x $
$\Rightarrow I =\int\limits_{0}^{\pi / 4} \log (1+\tan x) d x $
$=\int\limits_{0}^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x $
$=\int\limits_{0}^{\pi / 4} \log \left(1+\frac{\tan \pi / 4-\tan x}{1+\tan \pi / 4 \tan x}\right) d x $
$=\int\limits_{0}^{\pi} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x $
$\Rightarrow I =\int\limits_{0}^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x$
On adding Eqs. (i) and (ii), we get
$2 I =\int\limits_{0}^{\pi / 4} \log 2 d x=\log 2\left(\frac{\pi}{4}\right) $
$\Rightarrow I=\frac{\pi}{8} \log \,2$