Question

найти интеграл dx/(sin^2*3x)

Answer 1

$intfrac{dx}{sin^23x}=left(begin{array}{c}u=ctg3x=frac{cos3x}{sin3x}\du=frac{-3sin^2x-3cos^2x}{sin^23x}dx=-frac3{sin^23x}dx\-frac13du=frac1{sin^23x}end{array}right)=int-frac13du=-frac13int du=\=-frac13u+C=-frac13ctg3x+C$

Answer 2

Благодарю