Задача 33296 ...
Условие
Решение
sin^34x=sin^24x*sin4x=(1-cos^24x)*sin4x;
∫ sin^34xdx= ∫ (1-cos^24x)*sin4xdx= ∫ sin4xdx- ∫ cos^24x*sin4xdx=
=(1/4)*(-cos4x)- ∫ cos^24xd(cos4x)/(-4)=
=(1/4)*(-cos4x)+(1/4)*(cos^34x)/3+C=
=(-1/4)*cos4x+(1/12)cos^34x +C
14.
cos^4(x/5)=(cos^2(x/5))^2=((1+cos(2x/5))/2)^2=
=(1/4)+(1/2)cos(2x/5)+(1/4)cos^2(2x/5)=
=(1/4)+(1/2)cos(2x/5)+(1/4)*(1+cos(4x/5))/2=
(1/4)+(1/8)+(1/2)cos(2x/5)+(1/8)cos(4x/5)
∫cos^4(x/5)dx= ∫ ((3/8)+(1/2)cos(2x/5)+(1/8)cos(4x/5))dx=
=(3/8)x+(5/4)sin(2x/5)+(5/32)sin(4x/5)+C
15.
sin^4(x/3)*cos^2(x/3)=(sin^2(x/3))^2*cos^2(x/3)
применяем формулы понижения степени:
(1-cos(2x/3))^2*(1+cos(2x/3)=
=(1-cos^2(2x/3))*(1-cos(2x/3))=sin^2(2x/3)*(1-cos(2x/3)
∫ sin^4(x/3)*cos^2(x/3)dx= ∫ sin^2(2x/3)*(1-cos(2x/3)dx=
= ∫sin^2(2x/3)dx - ∫ sin^2(2x/3)cos(2x/3)dx=
= ∫ (1-cos(4x/3))dx/2 -(3/2) ∫ sin^2(2x/3)d(sin(2x/3))=
=(1/2)x - (3/8)*sin(4x/3)-(3/2)*(sin^3(2x/3))/3 +C=
=(1/2)x - (3/8)*sin(4x/3)-(1/2)*(sin^3(2x/3)) +C
17.
ctg^34x=ctg4x*(ctg^24x)=ctg4x*((1/sin^2x)-1)
d(ctg4x)=(ctg4x)`dx=(-1/sin^24x)*(4x)`=-4dx/sin^24x
∫ ctg^34xdx= ∫ ctg4x*dx(1/sin^2x)- ∫ ctg4x=
=(-1/4) ∫ ctg4x d(ctg4x)- ∫ cos4xdx/sin4x=
=(-1/4)(ctg^2(4x))/2-(1/4) ∫ d(sin4x)/sin4x=
(-1/8)ctg^2(4x)-(1/4)ln|sin4x|+С
18.
3cos4x-2sin4x+1=3*(cos^22x-sin^22x)-2*2sin2xcos2x+sin^22x+cos^22x=
=4cos^22x-4sin2x*cos2x-2sin^22x= cos^22x*(4-4tg2x-2tg^22x)
tg2x=t
-2t^2-4t+4=-2(t^2+2t+1)+6=-2(t+1)^2+6=6-2(t+1)^2
d(tg2x)=(tg2x)`dx=(2x)`dx/cos^22x
dx/cos^2(2x) = dt/2
получим табличный интеграл
(1/4)∫dt/(3-(t+1)^2)=(1/4) *(1/(2sqrt(3)))ln | (sqrt(3)+tg2x+1)/(sqrt(3)-tg2x-1)| + C