Definite Integration Question 136
Question: $ \int_0^{1}{{{\sin }^{-1}}( \frac{2x}{1+x^{2}} )dx=} $
[Karnataka CET 1999]
Options:
A) $ \frac{\pi }{2}-2\log \sqrt{2} $
B) $ \frac{\pi }{2}+2\log \sqrt{2} $
C) $ \frac{\pi }{4}-\log \sqrt{2} $
D) $ \frac{\pi }{4}+\log \sqrt{2} $
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Answer:
Correct Answer: A
Solution:
Put $ x=\tan \theta , $
$ \therefore $ $ dx={{\sec }^{2}}\theta d\theta $
As $ x=1\Rightarrow \theta =\frac{\pi }{4} $ and $ x=0\Rightarrow \theta =0 $ , then $ I=2\int_0^{\pi /4}{\theta {{\sec }^{2}}\theta d\theta =2[\theta \tan \theta ]_0^{\pi /4}-2\int_0^{\pi /4}{\tan \theta d\theta }} $
= $ \frac{\pi }{2}+2[\log \cos x]_0^{\pi /4}=\frac{\pi }{2}-2\log \sqrt{2} $ .
BETA
