Question

The value of $\int _ { 0 } ^ { \pi / 4 } \frac { 1 + \tan x } { 1 - \tan x } d x$ is

• A. $- \frac { 1 } { 2 } \log 2$
• B. $\frac { 1 } { 4 } \log 2$
• C. $\frac { 1 } { 3 } \log 2$
• D. None of these

Answer 1

Answer: (A)

$- \frac { 1 } { 2 } \log 2$

Answer 2

$\int _ { 0 } ^ { \pi / 4 } \frac { 1 + \tan x } { 1 - \tan x } d x = \int _ { 0 } ^ { \pi / 4 } \tan \left( \frac { \pi } { 4 } + x \right) d x$

$= \left[ \log \left\{ \sec \left( \frac { \pi } { 4 } + x \right) \right\} \right] _ { 0 } ^ { - \pi / 4 } = - \frac { 1 } { 2 } \log 2$ .