Задача 54959 Длина дуги кривой у=sqrt (1–x2) + arccos...
Условие
Решение
[m] y`=\frac{1}{2\sqrt{1-x^2}}\cdot (-2x)`-\frac{1}{\sqrt{1-x^2}}[/m]
[m] y`=-\frac{x}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}[/m]
[m] y`=-\frac{(x+1)}{\sqrt{(1-x)(1+x)}}[/m]
[m] y`=-\sqrt{\frac{1+x}{1-x}}[/m]
[m] L=\int^{\frac{8}{9}}_{0}\sqrt{1+(-\sqrt{\frac{1+x}{1-x}})^2}dx=\int^{\frac{8}{9}}_{0}\sqrt{\frac{1-x+1+x}{1-x}}dx=\int^{\frac{8}{9}}_{0}\sqrt{\frac{2}{1-x}}dx=\sqrt{2}\int^{\frac{8}{9}}_{0}\frac{1}{\sqrt{1-x}}dx=[/m]
[m]=\sqrt{2}\cdot (-2\sqrt{1-x})|^{\frac{8}{9}}_{0}=-2\sqrt{2}\cdot (\sqrt{1-\frac{8}{9}}-\sqrt{1})=\frac{4\sqrt{2}}{3}[/m]
[m]\int \frac{dx}{\sqrt{1-x}}=-\int \frac{d(1-x)}{\sqrt{1-x}}=-2\sqrt{1-x}[/m] см формулу 5 в таблице 
