Задача 77743 ...
Условие
Решение
Универсальная тригонометрическая подстановка:
[m]t = tg \frac{x}{2}; \sin x = \frac{2t}{1+t^2}; \cos x = \frac{1-t^2}{1+t^2}; dx = \frac{2dt}{1+t^2}[/m]
[m]\int \frac{dx}{5+\sin x + 3\cos x} = \int \frac{2dt}{1+t^2} : (5+\frac{2t}{1+t^2} + \frac{3(1-t^2)}{1+t^2}) =[/m]
[m]= \int \frac{2dt}{1+t^2} : \frac{5+5t^2+2t+3 - 3t^2}{1+t^2} = \int \frac{2dt}{1+t^2} \cdot \frac{1+t^2}{2t^2+2t+8} = \int \frac{dt}{t^2+t+4}=[/m]
[m]= \int \frac{dt}{t^2+2t \cdot 1/2 + 1/4 - 1/4+4} =\int \frac{dt}{(t+1/2)^2 + 15/4} =\int \frac{dt}{(t+1/2)^2 + (\sqrt{15}/2)^2}= [/m]
[m]= \frac{2}{\sqrt{15}} arctg\ \frac{2(t+1/2)}{\sqrt{15}} + C = \frac{2}{\sqrt{15}} arctg\ \frac{2t+1}{\sqrt{15}} + C [/m]
Возвращаемся к переменной x:
[m]\int \frac{dx}{5+\sin x + 3\cos x} = \frac{2}{\sqrt{15}} arctg\ \frac{2tg\ (x/2) + 1}{\sqrt{15}} + C[/m]