Question

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

• A. $\pi log2$
• B. $\frac{\pi }{8}log2$
• C. $\frac{\pi }{4}log2$
• D. $-\pi log2$

Answer 1

Answer: (B)

$\frac{\pi }{8}log2$

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

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

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

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

$I={\int }_0^{\frac{\pi }{4}}log2-I$

$2I=\frac{\pi }{4}log2$

Thus $I=\frac{\pi }{8}log2$