Предмет: Математика,
автор: Kasper93
Решите пожалуйста!
Найдите первообразную функцию:
1) 
2)
Ответы
Автор ответа:
0
1) f(x) = 1/x^2 = x^(-2); F(x) = x^(-1)/(-1) = -1/x + C
2) f(x) = sin(x/3)*cos x
F(x) = Int sin(x/3)*cos x dx
u = sin(x/3); dv = cos x dx; du = 1/3*cos(x/3) dx; v = sin x
F(x) = sin(x/3)*sin x - 1/3*Int cos(x/3)*sin x dx
u = cos(x/3) dx; dv = sin x dx; du = 1/3*(-sin(x/3)) dx; v = -cos x
F(x) = sin(x/3)*sin x - 1/3*(-cos(x/3)*cos x - 1/3*Int sin(x/3)*cos x dx) =
= sin(x/3)*sin x + 1/3*cos(x/3)*cos x + 1/9*Int sin(x/3)*cos x dx
F(x) = sin(x/3)*sin x + 1/3*cos(x/3)*cos x + 1/9*F(x)
8/9*F(x) = sin(x/3)*sin x + 1/3*cos(x/3)*cos x
F(x) = 9/8*(sin(x/3)*sin x + 1/3*cos(x/3)*cos x) + C
2) f(x) = sin(x/3)*cos x
F(x) = Int sin(x/3)*cos x dx
u = sin(x/3); dv = cos x dx; du = 1/3*cos(x/3) dx; v = sin x
F(x) = sin(x/3)*sin x - 1/3*Int cos(x/3)*sin x dx
u = cos(x/3) dx; dv = sin x dx; du = 1/3*(-sin(x/3)) dx; v = -cos x
F(x) = sin(x/3)*sin x - 1/3*(-cos(x/3)*cos x - 1/3*Int sin(x/3)*cos x dx) =
= sin(x/3)*sin x + 1/3*cos(x/3)*cos x + 1/9*Int sin(x/3)*cos x dx
F(x) = sin(x/3)*sin x + 1/3*cos(x/3)*cos x + 1/9*F(x)
8/9*F(x) = sin(x/3)*sin x + 1/3*cos(x/3)*cos x
F(x) = 9/8*(sin(x/3)*sin x + 1/3*cos(x/3)*cos x) + C
Похожие вопросы
Предмет: Русский язык,
автор: Motovilova02
Предмет: Математика,
автор: 006batir
Предмет: Литература,
автор: illaluniv236
Предмет: Биология,
автор: muratowaasya
Предмет: Литература,
автор: Янайоу