Question

Решить определенный интеграл подробно

Answer 1

$\displaystyle\int\limits^{\frac{\pi}{2}}_0\frac{cosx}{5+4cosx}dx=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_0\frac{5+4cosx-5}{5+4cosx}dx=\\=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_0dx-\frac{5}{4}\int\limits^{\frac{\pi}{2}}_0\frac{dx}{5+4cosx}=\\=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_0dx-\frac{5}{2}\int\limits^1_0\frac{1}{1+t^2}\frac{1+t^2}{t^2+9}dt=\\=\frac{1}{4}\int\limits^{\frac{\pi}{2}}_0dx-\frac{5}{2}\int\limits^1_0\frac{dt}{t^2+9}=\frac{x}{4}|^\frac{\pi}{2}_0-\frac{5}{6}(arctg\frac{t}{3})|^1_0=$

$\displaystyle=\frac{\pi}{8}-\frac{5}{6}arctg\frac{1}{3}\\\\\\t=tg\frac{x}{2};x=2arctgt;dx=\frac{2dt}{1+t^2}\\t_1=tg\frac{\pi}{4}=1;t_2=tg0=0\\5+4cosx=5+4\frac{1-t^2}{1+t^2}=\frac{9+t^2}{1+t^2}$