Задача 53458 ...
Условие
Найдите неопределенный интеграл
Sx*cos2xdx
Sx*(2x-1)^7dx
Sx*√x-3 dx
Решение
[i] по частям[/i]
u=x
dv=cos2xdx
du=dx
v= ∫ dv= ∫ cos2xdx=(1/2) ∫ cos[b]2x[/b]d([b]2x[/b])=(1/2)*[b](sin2x)[/b]
∫ u*dv=u*v- ∫ v*du
∫ x*cos2xdx=x*(1/2)*(sin2x)- ∫ (1/2)*(sin2x)dx=(1/2)*x*sin2x-(1/2) ∫ sin2xdx=
=(1/2)x*sin2x-(1/2)*(1/2) ∫ sin[b]2x[/b]d([b]2x[/b])=(1/2)*x*sin2x-(1/4)(-cos2x)+C[b]=(1/2)*x*sin2x+(1/4)*cos2x+C[/b]
2)
∫ x·(2x–1)^7dx= [i]замена переменной:[/i]
2x-1=t
2x=t+1
x=(t+1)/2
dx=(1/2)dt
= ∫ (t+1)/2 *t^2*(1/2)dt= (1/4)∫ (t+1)*t^7dt= (1/4)∫ (t^8+t^7)dt=(1/4)*(t^9/9)+(1/4)*(t^8/8)+C=
=(1/36)*(2x-1)^9 +(1/32)*(2x-1)^8 + C
3)
∫ x*sqrt(x-3)dx= [i]замена переменной:[/i]
sqrt(x-3)=t
x-3=t^2
x=t^2+3
dx=2tdt
∫ (t^2+3)*t*2tdt= ∫ (2t^4+6t^2)dt= 2*(t^5/5)+6*(t^3/3)+C=
=(2/5)t^5+2t^3+C=
=(2/5)*(sqrt(x-3))^5+2*(sqrt(x-3))^3+C=
=(2/5)(x-3)^2*sqrt(x-3)+2(x-3)*sqrt(x-3) +C.