Question

$\int _ { 0 } ^ { \pi / 2 } \log \tan x d x =$

• A. $\frac { \pi } { 2 } \log _ { e } 2$
• B. $- \frac { \pi } { 2 } \log _ { e } 2$
• C. $\pi \log _ { e } 2$
• D. 0

Answer 1

Answer: (D)

0

Answer 2

$\int _ { 0 } ^ { \pi / 2 } \log \tan x d x = \int _ { 0 } ^ { \pi / 2 } \log \left( \frac { \sin x } { \cos x } \right) d x$

$= \int _ { 0 } ^ { \pi / 2 } \log \sin x d x - \int _ { 0 } ^ { \pi / 2 } \log \cos x d x = 0$,

$\left\{ \because \int _ { 0 } ^ { a } f ( x ) d x = \int _ { 0 } ^ { a } f ( a - x ) d x \right\}$ .