A) \[0\]
B) \[\pi \]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{4}\]
E) \[-\pi \]
Correct Answer: C
Solution :
Let \[I=\int_{0}^{1}{\frac{d}{dx}}\left[ {{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) \right]dx\] Let \[x=tan\theta \] \[\therefore \] \[I=\int_{0}^{1}{\frac{d}{dx}}[{{\sin }^{-1}}(\sin 2\theta )]dx\] \[=2\int_{0}^{1}{\frac{d}{dx}}[{{\tan }^{-1}}x]\,dx\] \[=2[{{\tan }^{-1}}(1)-{{\tan }^{-1}}(0)]\] \[=2.\frac{\pi }{4}=\frac{\pi }{2}\]
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