Question

$\int _ { 0 } ^ { 1 } \tan ^ { - 1 } x d x =$

• A. $\frac { \pi } { 4 } - \frac { 1 } { 2 } \log 2$
• B. $\pi - \frac { 1 } { 2 } \log 2$
• C. $\frac { \pi } { 4 } - \log 2$
• D. $\pi - \log 2$

Answer 1

Answer: (A)

$\frac { \pi } { 4 } - \frac { 1 } { 2 } \log 2$

Answer 2

Put $x = \tan \theta \Rightarrow d x = \sec ^ { 2 } \theta d \theta$

Also as $x = 0 , \theta = 0$and $x = 1 , \theta = \frac { \pi } { 4 }$

Therefore, $\int _ { 0 } ^ { 1 } \tan ^ { - 1 } x d x = \int _ { 0 } ^ { \pi / 4 } \theta \sec ^ { 2 } \theta d \theta$

$= \frac { \pi } { 4 }$ $- \log \sqrt { 2 } = \frac { \pi } { 4 } - \frac { 1 } { 2 } \log 2$ .