Question

над интегралом п/3 внизу п/6 ∫(1+ctg^2 x) dx

Answer 1

$\int _{\frac{\pi}{6}}^{\frac{\pi}{3}}(1+ctg^2x)dx=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{sin^2x}=-ctgx|_{\frac{\pi}{6}}^{\frac{\pi}{3}}=-(ctg\frac{\pi}{3}-ctg\frac{\pi}{6})=\\\\=-(\frac{\sqrt3}{3}-\sqrt3)=\frac{2\sqrt3}{3}$