Question

Evaluate ${\int }_0^{\frac{\pi }{4}}\log (1+\tan x)dx$

• A. $\frac{\pi }{8}\log 2$
• B. $\frac{3\pi }{8}\log 2$
• C. $\frac{\pi }{4}\log 2$
• D. $\frac{3\pi }{4}\log 2$

Answer 1

The correct option is A $\frac{\pi }{8}\log 2$

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

$\Rightarrow I={\int }_0^{\frac{\pi }{4}}\log (1+\tan (\frac{\pi }{4}-x))dx$.......[$\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$]

$={\int }_0^{\frac{\pi }{4}}\log (1+\frac{1-\tan x}{1+\tan x})dx$......[$\because \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$]

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

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

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