Question

The value of ${\int }_0^{2/3}\frac{dx}{4+9x^2}$ is

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

Answer 1

Answer: (D)

${\int }_0^{2/3}\frac{dx}{4+9x^2}={\int }_0^{2/3}\frac{dx}{9\left(\frac{4}{9}+x^2\right)}$
$=\frac{1}{9}{\int }_0^{2/3}\frac{dx}{{\left(\frac{2}{3}\right)}^2+x^2}$
$=\frac{1}{9}{\left[\frac{1}{2/3}{\tan }^{-1}\left(\frac{x}{2/3}\right)\right]}_0^{2/3}$
$=\frac{1}{9}\left[\frac{3}{2}{\tan }^{-1}\left(\frac{2/3}{2/3}\right)-\frac{3}{2}{\tan }^{-1}(0)\right]$
$=\frac{1}{9}\left[\frac{3}{2}{\tan }^{-1}\left(\tan \frac{\pi }{4}\right)-\frac{3}{2}\times 0\right]$
$=\frac{1}{3.2}\left(\frac{\pi }{4}\right)$
$=\frac{\pi }{24}$