## Calculus 10th Edition

$\frac{\pi}{18} \approx$ 0.175
$\int_0^\frac{2}{\sqrt 3}\frac{1}{{4+9x^{2}}}dx$ $=\frac{1}{3}\int_0^\frac{2}{\sqrt 3}\frac{1}{{2^{2}+(3x)^{2}}}dx$ Use arctan identity: where a = 2, u = 3x, and du = 3dx $\int{\frac{du}{a^{2}+u^{2}}} = \frac{1}{a}arctan(\frac{u}{a}) + C$ $=[\frac{1}{6}arctan(\frac{3x}{2})]^{\frac{2}{\sqrt 3}}_{0}$ $=[\frac{1}{6}arctan(\frac{6}{2\sqrt 3})]-[\frac{1}{6}arctan(0)]$ $=[(\frac{1}{6})(\frac{\pi}{3})]-0$ $=\frac{\pi}{18}$