Definite Integration Question 245
Question: $ \int_0^{1}{{{\tan }^{-1}}xdx=} $
[Karnataka CET 1993; RPET 1997]
Options:
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 $
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta d\theta $
Also as $ x=0,\theta =0 $ and $ x=1,\theta =\frac{\pi }{4} $
Therefore, $ \int_0^{1}{{{\tan }^{-1}}xdx=\int_0^{\pi /4}{\theta {{\sec }^{2}}\theta d\theta }} $
$ =\frac{\pi }{4} $ $ -\log \sqrt{2}=\frac{\pi }{4}-\frac{1}{2}\log 2 $ .
BETA
