Question

помогите пожалуйста срочно!

Answer 1

$\int\limits_\frac{\pi}{6}^\frac{\pi}{3}\frac{dx}{cos^2x}=tgx|^\frac{\pi}{3}_\frac{\pi}{6}=tg(\frac{\pi}{3})-tg(\frac{\pi}{6})=\sqrt{3}-\frac{1}{\sqrt{3}}=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$



$\int\limits^1_0(0.4\sqrt{x^3+1}x^2)dx=0.4\int\limits^1_0\sqrt{x^3+1}*x^2*\frac{d(x^3+1)}{3x^2}=\\=\frac{2}{15}\int\limits^1_0(x^3+1)^\frac{1}{2}d(x^3+1)=\frac{2}{15}*(\frac{(x^3+1)^\frac{3}{2}}{\frac{3}{2}})|^1_0=\\=\frac{2}{15}*\frac{2}{3}((1^3+1)^{\frac{3}{2}}}{}-(0^3+1)^\frac{3}{2})=\frac{4}{45}(2\sqrt{2}-1)$