Question

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

Answer 1

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

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

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

$=frac{sqrt2}{6}arctgfrac{ctg x}{sqrt2}+C$