Question

решить неопределенный интеграл

Answer 1

$\displaystyle\int\frac{dx}{5+2sinx+3cosx}=\int\frac{dt}{t^2+2t+4}=\int\frac{d(t+1)}{(t+1)^2+3}=\\=\frac{1}{\sqrt3}arctg\frac{t+1}{\sqrt3}+C=\frac{1}{\sqrt3}arctg\frac{tg\frac{x}{2}+1}{\sqrt3}+C\\t=tg\frac{x}{2};x=2arctgt\\dx=\frac{2dt}{1+t^2}\\5+2sinx+3cosx=5+\frac{4t}{1+t^2}+\frac{3-3t^2}{1+t^2}=\frac{2(t^2+2t+4)}{1+t^2}$