Question

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

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{24}$
• D. $\frac{\pi }{18}$

Answer 1

The correct option is C $\frac{\pi }{24}$

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

Put $t=\frac{3}{2}x\Rightarrow dt=\frac{3}{2}dx$ or $dx=\frac{2}{3}dt$

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

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

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

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

$=\frac{1}{6}[\frac{\pi }{4}-0]=\frac{\pi }{24}$