Задача 35330 ...
Условие
Решение
∫ сos3x*cos2xdx
Формула
сos α *cos β =(1/2)cos( α + β )+(1/2)cos( α - β )
сos3x*cos2x=(1/2)cos5x+(1/2)cosx
∫ сos3x*cos2xdx= ∫ ((1/2)cos5x+(1/2)cosx)dx
интеграл от суммы равен сумме интегралов:
=(1/2) ∫ сos5xdx+(1/2) ∫ cosxdx=(1/2)*(1/5)*sin5x +(1/2)sinx+C=
= [b](1/10)sin5x+(1/2)sinx+C[/b]
6.
cos^62x=(cos^22x)^3=((1+cos4x)/2)^3=(1+3cos4x+3cos^24x+cos^34x)/8
Интеграл от суммы равен сумме интегралов:
∫ cos^62xdx=(1/8)∫ dx+(3/8)∫ cos4xdx+(3/8)∫ cos^24xdx+(1/8)∫ cos^34xdx=
=(1/8)x+(3/8)*(1/4)*sin4x +(3/8)*(1/2) ∫ (1+cos8x)dx+(1/8) ∫ (1-sin^24x)*cos4xdx=
= [b](1/8)x[/b] +(3/32)sin4x+ [b](3/16)x[/b] +(3/16)*(1/8)*sin8x+
+(1/8)*(1/4)*sin4x-(1/8)*(1/4)*(sin^34x/3)+C=
привести подобные
5.
∫ sin^4x*cos^3xdx= ∫ sin^4x*cos^2x*cosxdx=
= ∫ sin^4x*(1-sin^2x)*cosxdx= ∫ sin^4x*cosxdx- ∫ sin^6xcosxdx=
= ∫ sin^4x*d(sinx)- ∫ sin^6xd(sinx) [b]=(sin^5x/5)-(sin^7x/7) + C[/b]
3.
2sin^2x+7cos^2x=cos^2x*(2tg^2x+7)
∫ dx/(2sin^2x+7cos^2x)= ∫dx/ cos^2x(2tg^2x+7)=
=(1/2) ∫ d(tgx)/(tg^2x+(7/2)= (1/2)*(1/sqrt(7/2))arctg(tgx/sqrt(7/2)C=
= [b](1/sqrt(14)) arctg (sqrt(2)tgx/sqrt(7)) + C[/b]
1.
tg(x/2)=t
x/2=arctgt
x=2arctgt
dx=2dt/(1+t^2)
cosx=(1-t^2)/(1+t^2)
2-cosx=1-(1-t^2)/(1+t^2)=(2+2t^2-1+t^2)/(1+t^2)=(3t^2+1)/(t^2+1)
∫ dx/(2-cosx)= ∫ dt/(3t^2+1)= (1/3) ∫ dt/(t^2+(1/3))=
=(1/3)*(1/sqrt(1/3))* arctg (t/sqrt(1/3))+C=
= [b](1/sqrt(3)) arctg(sqrt(3)*tg(x/2)) + C[/b]
2.
cos^5x=cos^4x*cosx=(cos^2x)^2*cosx=(1-sin^2x)^2*cosx=
=(1-sinx)^2*(1+sinx)^2*cosx
∫ cos^5xdx/(1+sinx)= ∫ (1-sinx)^2*(1+sinx)cosxdx
sinx=t
cosxdx=dt
= ∫ (1-t^2)*(1+t)dt= ∫ (1-t^2+t-t^3)dt= t-(t^3/3)+(t^2/2)-(t^4/4) + C=
= [b]sinx- (sinx)^3/3 + (sinx)^2/2 - (sinx)^4/4 + C[/b]
4.
tgx=t
x=arctgt
dx=dt/(1+t^2)
1+tg^2t=1/cos^2x
cos^2x=1/(1+tg^2x)=1/(1+t^2)
sin^2x=1-cos^2x=1-(1/(1+tg^2x))=tg^2x/(1+tg^2x)=t^2/(1+t^2)
∫ sin^2xdx/cos^(10)x=
= ∫ t^2*(1+t^2)^5dt/(1+t^2)^2=
= ∫ t^2*(1+t^2)^3dt=
= ∫ (t^2+3t^4+3t^6+t^8)dt=
=(t^3/3)+(3t^5/5)+(3t^7/7)+(t^9/9) + C=
= [b](tg^3x/3)+(3tg^5x/5)+(3tg^7x/7)+(tg^9x/9) + C
[/b]
