Question

Вычислить определенный интеграл

Answer 1

$\int\limits_6^{6\sqrt{3}}\frac{dx }{x^2+36}=\frac{1}{\sqrt{36}}\arctan \frac{x}{\sqrt{36}}|_6^{6\sqrt{3}}=$

$=\frac{1}{6}(\arctan\frac{6\sqrt{3}}{6}-\arctan\frac{6}{6})=\frac{1}{6}(\arctan\sqrt{3}-\arctan1)=$

$=\frac{1}{6}(\frac{\pi}{3}-\frac{\pi}{4})=\frac{1}{6}\frac{\pi}{12}=\frac{\pi}{72}$

Ответ: $\frac{\pi}{72}$