Задача 78848 [m]\int_{0}^{1/2} \frac{8x - arctg...
Условие
[m]\int_{0}^{1/2} \frac{8x - arctg 2x}{1 + 4x^2}dx[/m]
Решение
1 интеграл берем заменой: 1 + 4x^2 = y; dy = 8x dx; y(0) = 1; y(1/2) = 1 + 4*1/4 = 2
2 интеграл берем заменой: arctg 2x = t; dt = dx/(1 + 4x^2); t(0) = 0; t(1/2) = arctg 1
[m]\int \limits_0^{1/2} \frac{8x}{1 + 4x^2}\ dx - \int \limits_0^{1/2} \frac{arctg\ 2x}{1 + 4x^2}\ dx = \int \limits_1^2 \frac{dy}{y} - \int \limits_0^{arctg\ 1} t\ dt =[/m]
[m]= \ln |y| \Big |_1^2 - \frac{t^2}{2} \Big |_0^{arctg\ 1} = \ln 2 - \ln 1 - \frac{arctg^2\ 1}{2}+ \frac{arctg^2\ 0}{2} = \ln 2 - \frac{arctg^2\ 1}{2}[/m]