Question

Решить определенный интеграл подробно

Answer 1

$displaystyleintlimits^{frac{pi}{2}}_0frac{cosx}{5+4cosx}dx=frac{1}{4}intlimits^{frac{pi}{2}}_0frac{5+4cosx-5}{5+4cosx}dx=\=frac{1}{4}intlimits^{frac{pi}{2}}_0dx-frac{5}{4}intlimits^{frac{pi}{2}}_0frac{dx}{5+4cosx}=\=frac{1}{4}intlimits^{frac{pi}{2}}_0dx-frac{5}{2}intlimits^1_0frac{1}{1+t^2}frac{1+t^2}{t^2+9}dt=\=frac{1}{4}intlimits^{frac{pi}{2}}_0dx-frac{5}{2}intlimits^1_0frac{dt}{t^2+9}=frac{x}{4}|^frac{pi}{2}_0-frac{5}{6}(arctgfrac{t}{3})|^1_0=$

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