Задача 71093 integrate (3x + 1) * cos 2x dx...

644a824f6782b137f1d3645b

27.04.2023 21:30:03

integrate (3x + 1) * cos 2x dx

5f3ea7e3faf909182968ddd9

28.04.2023 08:00:47

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По частям

u=[b]3x+1[/b]

du=[u]3dx[/u]

dv=cos2x dx

v=[blue](1/2)sin2x[/blue]


∫ (3x+1)*cos2xdx=[b](3x+1)[/b]*(1/2)*sin2x - ∫[blue] (1/2)sin2x[/blue] * [u]3dx[/u]=(1/2)*(3x+1)*cos2x-(3/2)*(1/2)*(-cos2x)+C