Sin2*x-cos*3x+1 решить определенный интеграл (1; -1) - вопрос №231119186716 от chuvak1414 24.10.2021 12:01

Question

Алгебра: Sin2*x-cos*3x+1 решить определенный интеграл (1; -1)

chuvak1414 · Answer

[-1;1] ∫ (sin2x-cos3x+1) dx =
= [-1;1] (-1/2 cos2x - 1/3 sin3x +x) =
= -1/2 ( cos (2*1) - cos(2*(-1) ) - 1/3 ( sin(3*1) - sin(3*(-1) ) + 1 -(-1) =
= -1/2 ( cos (2) - cos(-2) ) - 1/3 ( sin(3) - sin(-3) ) + 2 =
= -1/2 ( cos (2) - cos(2) ) - 1/3 ( sin(3) + sin(3) ) + 2 =
= -1/2 * 0 - 1/3 * 2sin(3) + 2 =
= 2/3 * ( 3 - sin(3) )