1. Вычислите интеграл: 1) \\[\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{dx}{cos^2 x}; \] 2) \\[\int_{1}^{2} (6x^2 + \frac{1}{x^2})dx.\\]

1) \\[\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{dx}{cos^2 x} = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (tan(x))' dx = tan(x)|_{\frac{\pi}{6}}^{\frac{\pi}{4}} = tan(\frac{\pi}{4}) - tan(\frac{\pi}{6}) = 1 - \frac{\sqrt{3}}{3} \]

2) \[\int_{1}^{2} (6x^2 + \frac{1}{x^2})dx = \int_{1}^{2} 6x^2 dx + \int_{1}^{2} \frac{1}{x^2} dx = 6 \int_{1}^{2} x^2 dx + \int_{1}^{2} x^{-2} dx = 6 \cdot \frac{x^3}{3} |_{1}^{2} + \frac{x^{-1}}{-1} |_{1}^{2} = 2x^3 |_{1}^{2} - \frac{1}{x} |_{1}^{2} = 2(2^3 - 1^3) - (\frac{1}{2} - \frac{1}{1}) = 2(8-1) - (\frac{1}{2} - 1) = 2 \cdot 7 - (-\frac{1}{2}) = 14 + \frac{1}{2} = 14.5 \\]