A) \[\frac{\pi }{12}\]
B) \[\frac{\pi }{24}\]
C) \[\frac{\pi }{4}\]
D) 0
Correct Answer: B
Solution :
\[\int_{0}^{2/3}{\frac{dx}{4+9{{x}^{2}}}=\frac{1}{9}\int_{0}^{2/3}{\frac{dx}{{{(2/3)}^{2}}+{{x}^{2}}}}}\] \[=\frac{1}{9}\times \frac{1}{2/3}\left( {{\tan }^{-1}}\frac{x}{2/3} \right)_{0}^{2/3}=\frac{\pi }{4}\times \frac{1}{6}=\frac{\pi }{24}\].
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