Question

The value of $ \int_{0}^{2/3}{\frac{dx}{4+9{{x}^{2}}}} $ is

• A. $ \frac{\pi }{4} $
• B. $ \pi $
• C. $ 0 $
• D. $ \frac{\pi }{24} $

Answer 1

Answer: (D)

$ \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} $