Question

dx/3-2sin^2x найти интеграл

Answer 1

$displaystyleintfrac{dx}{3-2sin^2x}=intfrac{dx}{sin^2x(frac{3}{sin^2x}-2)}=-frac{1}{sqrt3}intfrac{d(sqrt3ctgx)}{3ctg^2x+1}=\=-frac{1}{sqrt3}arctg(sqrt{3}ctgx)+C\\\frac{3}{sin^2x}-2=3ctg^2x+1$