[ {{\sin }^{-1}}( \frac{2x}{1+x^{2}} ) ],dx} $ is equal to [Kerala (Engg.) 2002]
A) 0
B) $ \pi $
C) $ \pi /2 $
D) $ \pi /4 $
Correct Answer: C
$ I=[ {{\sin }^{-1}}( \frac{2x}{1+x^{2}} ) ]_0^{1}={{\sin }^{-1}}(1)-{{\sin }^{-1}}(0) $ $ =\frac{\pi }{2} $ .
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