A) 0
B) \[\pi \]
C) \[\pi /2\]
D) \[\pi /4\]
Correct Answer: C
Solution :
\[I=\left[ {{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) \right]_{0}^{1}={{\sin }^{-1}}(1)-{{\sin }^{-1}}(0)\]\[=\frac{\pi }{2}\].You need to login to perform this action.
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