Question

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

Answer 1

${\int }_0^{2/3}\frac{1}{4+9x^2}dx$

$=\frac{1}{9}$${\int }_0^{2/3}\frac{1}{(2/3{)}^2+x^2}dx$

$\left[\frac{1}{6}{\tan }^{-1}\right(\frac{x}{2/3}){]}_0^{2/3}$

$=\frac{1}{6}{\tan }^{-1}\left(1\right)$

$=\frac{\pi }{24}$