Question

Evaluate ${\int }_0^{2/3}\frac{dx}{4+{9x}^2}$

Answer 1

${\int }_0^{\frac{2}{3}}\frac{dx}{4+9x^2}=\frac{1}{9}{\int }_0^{\frac{2}{3}}\frac{dx}{\frac{4}{9}+x^2}=\frac{1}{9\times \frac{2}{3}}\times {\left[{\tan }^{-1}\frac{x}{\frac{2}{3}}\right]}_0^{\frac{2}{3}}=\frac{1}{6}\times \left[\frac{\pi }{4}\right]=\frac{\pi }{24}$