Question

найти интеграл sin^4*3xdx

Answer 1

sin ^ 4 3xdx = (1/3) ∫sin ^ 4 уду, и = 3x = (1/3) ∫ [(1-cos2u) / 2] ^ 2 Udu = (1/12) ∫1-2cos2u + ( 1/2) (1 + cos4u) ди = (1/12) [и - грех 2u + (1/2) и + (1/8) sin4u] + с = (3/8) х - (1/12 ) sin6x + (1/96) sin12x Вроде так

Answer 2

+С