Question

$\underset{0}{\overset{\pi /4}{\int }}\log (\frac{\sin x+\cos x}{\cos x})dx$ 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)

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