Question

Найти интеграл dx/(sin^2*3x)

Answer 1

$\int\frac{dx}{\sin^23x}=\left(\begin{array}{c}u=ctg3x=\frac{\cos3x}{\sin3x}\\du=\frac{-3\sin^2x-3\cos^2x}{\sin^23x}dx=-\frac3{\sin^23x}dx\\-\frac13du=\frac1{\sin^23x}\end{array}\right)=\int-\frac13du=-\frac13\int du=\\=-\frac13u+C=-\frac13ctg3x+C$