Задача 49368 ...

sova

2020-05-05 20:48:57

[m]d(arctg2x)=(arctg2x)`dx=\frac{1}{1+(2x)^2}\cdot (2x)`dx=\frac{1}{1+4x^2}\cdot 2=[/m][m]=\frac{2}{1+4x^2}dx[/m] ⇒

[m]\frac{1}{4+x^2}dx=\frac{1}{2}d(arctg2x)[/m]


∫ ^(+ ∞ )_(0)[m](arctg2x)^{\frac{1}{2}})\cdot \frac{1}{2}d(arctg2x)=[/m]

[m]= \frac{1}{2}\cdot \frac{(arctg2x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}[/m]|^(+ ∞ )_(0)=

[m]=\frac{1}{3}(arctg2x)^{\frac{3}{2}}[/m]|^(+ ∞ )_(0)=


[m]=\frac{1}{3}(\frac{\pi}{2})^{\frac{3}{2}}-0=[/m]

[m]=\frac{\pi}{6}\cdot \sqrt{\frac{\pi}{2}}[/m]