A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{2}\]
D) 0
E) \[\frac{\pi }{3}\]
Correct Answer: E
Solution :
Let \[I=\int_{-1/2}^{1/2}{\frac{dx}{{{(1-{{x}^{2}})}^{1/2}}}}\] Again let\[f(x)=\frac{1}{{{(1-{{x}^{2}})}^{1/2}}}\] \[f(-x)=\frac{1}{{{(1-{{x}^{2}})}^{1/2}}}=f(x)\] \[\therefore \] \[I=2\int_{0}^{1/2}{\frac{dx}{{{(1-{{x}^{2}})}^{1/2}}}}\] \[=\left( 2{{\sin }^{-1}}\frac{x}{1} \right)_{0}^{1/2}\] \[=2{{\sin }^{-1}}\frac{1}{2}=2\times \frac{\pi }{6}=\frac{\pi }{3}\]
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