найти интеграл от dx / (cos^2x-2sin^2x-4)

$\int{\frac{1}{cos^2x-2sin^2x-4}}\, dx=\int{\frac{1}{1-sin^2x-2sin^2x-4}}\, dx=$

$=\int{\frac{1}{-3sin^2x-3}}\, dx=-\frac{1}{3}\int{\frac{1}{sin^2x+1}}\, dx=-\frac{1}{3}\int{\frac{1}{sin^2x(1+\frac{1}{sim^2x})}}\, dx=$

$=\frac{1}{3}\int{\frac{1}{2+ctg^2x}}\, d(ctg x)=\frac{1}{3}*\frac{\sqrt2}{2}arctg\frac{ctg\ x}{\sqrt2}+C=$

$=\frac{\sqrt2}{6}arctg\frac{ctg\ x}{\sqrt2}+C$