ПОМОГИТЕ!НАЙДИТЕ ИНТЕГРАЛ!

Решите задачу:

$\int \frac{dx}{5-3cosx}= [\, t=tg\frac{x}{2},\; cosx= \frac{1-t^2}{1+t^2} ,\; dx= \frac{2dt}{1+t^2}\, ]=\\\\=\int \frac{ \frac{2dt}{1+t^2} }{5-3\cdot \frac{1-t^2}{1+t^2}} =\int \frac{2dt}{5(1+t^2)-3(1-t^2)} =\int \frac{dt}{4t^2+1}=\frac{1}{4}\int \frac{dt}{t^2+(\frac{1}{2})^2}=\\\\=\frac{1}{4}\cdot \frac{1}{\frac{1}{2}}}arctg\frac{t}{\frac{1}{2}}+C=\frac{1}{2}\cdot arctg(2tg\frac{x}{2})+C$