Решите 3(б) пожалуйста
∫ sin[b]u[/b] d[b]u[/b]=- cosu+C u=x/2 du=1/2dx ∫ sin(x/2)dx=2 ∫ sin(x/2)*(1/2)dx=2 ∫ sin(x/2)d(x/2)=-2cos(x/2) ∫ ^(3π/2)_(π/2)sin(x/2)dx=-2cos(x/2)|(3π/2)_(π/2)= =-2cos(3π/4)+2cos(π/4)=+2sqrt(2)/2 +2sqrt(2)/2=[b]2sqrt(2)[/b]