Интеграл (1-sin^3 x)/sin^2 x

Question

Интеграл (1-sin^3 x)/sin^2 x

Answer 1

∫(1-sin³x)/sin²x dx=∫dx/sin²x - ∫sin³x/sin²x dx=-ctgx-∫sinx dx=
=-ctgx-cosx + c
sinx≠0, x≠kπ,k∈Z