1)
sin(x)*sin(3x)
так как
sin (3x)= sin(2x + x) = sin(2x) cos(x) + sin(x)cos(2x), то
sin(x)*sin(3x)=sin(x)*[ sin(2x) cos(x) + sin(x)cos(2x)]=
=sin(x)*[2sin(x)cos(x)*cos(x)+sin(x)*(2cos^2(x)-1)]=
=sin^2(x)*[2cos^2(x)+2cos^2(x)-1]=sin^2(x)*[4cos^2(x)-1]=
=4sin^2(x)cos^2(x)-sin^2(x)
a. int(4sin^2(x)cos^2(x))dx=int(2sin(x)cos(x))^2dx=int(sin(2x)^2dx=
=int((1/2)*(1-cos(2*2x)))dx=(1/2)*(x-(1/4)*sin(4x))+c
б. int(sin^2(x))dx=(-1/2)int(1-cos(2x))dx=(-1/2)*[x-(1/2)sin(2x))]+c
итого
int sin(x)*sin(3x)dx=(1/2)*[x-(1/4)*sin(4x)]+c1+(-1/2)*[x-(1/2)sin(2x)]+c2=
=(1/2)*[(1/2)sin(2x)-(1/4)sin(4x)]+c
1)
sin(x)*sin(3x)
так как
sin (3x)= sin(2x + x) = sin(2x) cos(x) + sin(x)cos(2x), то
sin(x)*sin(3x)=sin(x)*[ sin(2x) cos(x) + sin(x)cos(2x)]=
=sin(x)*[2sin(x)cos(x)*cos(x)+sin(x)*(2cos^2(x)-1)]=
=sin^2(x)*[2cos^2(x)+2cos^2(x)-1]=sin^2(x)*[4cos^2(x)-1]=
=4sin^2(x)cos^2(x)-sin^2(x)
a. int(4sin^2(x)cos^2(x))dx=int(2sin(x)cos(x))^2dx=int(sin(2x)^2dx=
=int((1/2)*(1-cos(2*2x)))dx=(1/2)*(x-(1/4)*sin(4x))+c
б. int(sin^2(x))dx=(-1/2)int(1-cos(2x))dx=(-1/2)*[x-(1/2)sin(2x))]+c
итого
int sin(x)*sin(3x)dx=(1/2)*[x-(1/4)*sin(4x)]+c1+(-1/2)*[x-(1/2)sin(2x)]+c2=
=(1/2)*[(1/2)sin(2x)-(1/4)sin(4x)]+c
(130-х)дет. - выточил ученик
х/3=4(130-х)
х=12(130-х)
х=1560-12х
13х=1560
х=120(дет) - выточил токарь
130-120=10(дет). - выточил ученик