∫ sin^4x×cos^4x dx
∫ tg^4 x/2 dx
sin^4x*cos^4x=(1/16)sin^42x=(1/16)*(sin^22x)^2=(1/16)*((1-cos4x)/2)^2=
=(1/64)*(1-2cos4x+cos^24x)=(1/64)*(1-2cos4x+ (1+cos8x)/2)=
=(1/64)-(1/32)cos4x +(1/128)+(1/128)cos8x=
=(3/128)-(1/32)cos4x+(1/128)cos8x
∫ sin^4x*cos^4x dx= (3/128) ∫ dx - (1/32) ∫ cos4xdx+(1/128) ∫ cos8xdx=
=[b](3/128)x-(1/128)sin4x+(1/1024)sin8x+C[/b]
tg^4(x/2)=tg^2(x/2)*tg^2(x/2)=tg^2(x/2) *((1/cos^2(x/2)) -1)=
=tg^2(x/2)*(1/cos^2x/2) - tg^2(x/2)=
=tg^2(x/2)*(1/cos^2x/2) - ((1/cos^2(x/2)) -1)=
=tg^2(x/2)*(1/cos^2x/2) - (1/cos^2(x/2)) +1
∫ tg^4(x/2) dx= ∫ tg^2(x/2)*(1/cos^2x/2)dx - ∫ (1/cos^2(x/2))dx + ∫ dx=
= 2 ∫ tg^2(x/2) d(tg(x/2)) - 2 ∫ d(x/2)/cos^2(x/2) +x +c=
=2(tg^3(x/2))/3-2tg(x/2) + x + C=
=[b](2/3)*tg^3(x/2)-2tg(x/2) + x + C[/b]
так как
d(tg(x/2))=(1/cos^2(x/2))*(x/2)`dx ⇒
[blue]2d(tg(x/2)=dx/cos^2(x/2)[/blue]