Question

dx/3-2sin^2x найти интеграл

Answer 1

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