Решите пожалуйста!
Найдите первообразную функцию:
1) [tex]F(x)= \frac{1}{x^{2}} [/tex]

2) [tex]F(x)=sin \frac{x}{3}cos(x)[/tex]

1) f(x) = 1/x^2 = x^(-2); F(x) = x^(-1)/(-1) = -1/x + C2) f(x) = sin(x/3)*cos xF(x) = Int sin(x/3)*cos x dxu = sin(x/3); dv = cos x dx; du = 1/3*cos(x/3) dx; v = sin xF(x) = sin(x/3)*sin x - 1/3*Int cos(x/3)*sin x dxu = cos(x/3) dx; dv = sin x dx; du = 1/3*(-sin(x/3)) dx; v = -cos xF(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 dxF(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 xF(x) = 9/8*(sin(x/3)*sin x + 1/3*cos(x/3)*cos x) + C