A) \[\frac{\pi }{4}\]
B) \[\pi \]
C) \[0\]
D) \[\frac{\pi }{24}\]
Correct Answer: D
Solution :
\[\int_{0}^{2/3}{\frac{dx}{4+9{{x}^{2}}}}=\int_{0}^{2/3}{\frac{dx}{9\left( \frac{4}{9}+{{x}^{2}} \right)}}\] \[=\frac{1}{9}\int_{0}^{2/3}{\frac{dx}{{{\left( \frac{2}{3} \right)}^{2}}+{{x}^{2}}}}\] \[=\frac{1}{9}\left[ \frac{1}{2/3}{{\tan }^{-1}}\left( \frac{x}{2/3} \right) \right]_{0}^{2/3}\] \[=\frac{1}{9}\left[ \frac{3}{2}{{\tan }^{-1}}\left( \frac{2/3}{2/3} \right)-\frac{3}{2}{{\tan }^{-1}}(0) \right]\] \[=\frac{1}{9}\left[ \frac{3}{2}{{\tan }^{-1}}\left( \tan \frac{\pi }{4} \right)-\frac{3}{2}\times 0 \right]\] \[=\frac{1}{3.2}\left( \frac{\pi }{4} \right)\] \[=\frac{\pi }{24}\]
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