Question

Solve: ${\int }_0^{\pi /4}\log (1+\tan x)dx$.

• A. $\frac{\pi }{7}\log 2$
• B. $\frac{\pi }{8}\log 2$
• C. $-\frac{\pi }{8}\log 2$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi }{8}\log 2$

$I={\int }_0^{\pi /4}\log (1+\tan x)dx$

We know $I={\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$I={\int }_0^{\pi /4}\log (1+\tan x)dx={\int }_0^{\pi /4}\log (1+\tan (\frac{\pi }{4}-x))dx$

$I={\int }_0^{\pi /4}\log (1+\frac{\tan \pi /4-\tan x}{\tan \pi /4\tan x)+1})dx$

$={\int }_0^{\pi /4}\log \left(\frac{2}{1+\tan x}\right)dx$

$I={\int }_0^{\pi /4}\log 2-I$

$2I=\log 2\cdot \frac{\pi }{4}$

$I=\frac{\pi }{8}\log 2$.