Помогите решить интеграл

Универсальная тригонометрическая подстановка.

$\int \frac{dx}{5cosx+10sinx}=\Big [\, t=tg\frac{x}{2},\; sinx=\frac{2t}{1+t^2},\; cosx=\frac{1-t^2}{1+t^2},\; dx=\frac{2\, dt}{1+t^2}\, \Big ]=\\\\=\int \frac{2\, dt}{(1+t^2)\cdot (\frac{5(1-t^2)}{1+t^2}+\frac{20t}{1+t^2})}=\int \frac{2\, dt}{5-5t^2+20t}=-\frac{2}{5}\int \frac{dt}{t^2-4t-1}=\\\\=-\frac{2}{5}\int \frac{dt}{(t-2)^2-5}=-\frac{2}{5}\int \frac{d(t-2)}{(t-2)^2-5}=-\frac{2}{5}\cdot \frac{1}{2\sqrt5}\cdot ln\Big |\frac{t-2-\sqrt5}{t-2+\sqrt5}\Big |+C=\\\\=-\frac{1}{5\sqrt5}\cdot ln\Big |\frac{tg\frac{x}{2}-2-\sqrt5}{tg\frac{x}{2}-2+\sqrt5}\Big |+C$