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