Question

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

Answer 1

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

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

$=\frac{1}{4}{\int }_0^{\frac{2}{3}}\frac{dx}{1+{\left(\frac{3x}{2}\right)}^2}$

$=\frac{1}{4}\times \frac{2}{3}{\left[{\tan }^{-1}\frac{3x}{2}\right]}_0^{\frac{2}{3}}$

$=\frac{1}{6}[{\tan }^{-1}\frac{3\times 2}{2\times 3}-{\tan }^{-1}0]$

$=\frac{1}{6}[{\tan }^{-1}1-0]$

$=\frac{1}{6}\times \frac{\pi }{4}=\frac{\pi }{24}$