Задача 33349 Помогите 19 вариант, очень...
Условие
Неопределенный интеграл
Решение
Замена переменной
∛(2х+1)=t
2x+1=t^3
x=(t^3-1)/2
dx=3t^2dt/2
x+1=(t^3-1)/2 + 1= (t^3+1)/2
получаем
∫((t^3+1)/2)*(3t^2/2)dt/t= (3/4) ∫ (t^3+1)*tdt=(3/4) ∫ (t^4+t)dt=
=(3/4)*(t^5/5)+(3/4)*(t^2/2) + C
t=∛(2x+1)
=(3/20)∛(2x+1)^5 + (3/8)∛(2x+1)^2 + C
19б)
Замена
2+e^(2x)=t
e^(2x)=t-2
2x=ln(t-2)
x=(1/2)ln(t-2)
dx=(1/2)*(1/(t-2))dt
получаем
∫ (1/2)dt/(t*(t-2))=(1/2)*(1/2) ∫ (1/(t-2) - (1/t))dt=
=(1/4)ln| t -2| - (1/4) ln | t|+ C=
=(1/4) ln |e^(2x)| -(1/4) ln|2+e^(2x)|+C
= (1/4) ln (e^(2x)/(2+e^(2x)) + C
Дробь 1/(t*(t-2))= A/(t-2)+B/t A=1/2; B=-1/2
3
Тригонометрическая подстановка
x=5sint
25-x^2=25-25sin^2t=25*(1-sin^2t)=25cos^2t
dx=(5sint)`dt=5costdt
получим
∫( 5sint)^2*sqrt(25cos^2t)*(5costdt)=
=625 ∫ sin^2t*cos^2tdt= (формула sinx*cosx=(sin2x)/2)
=625/4 ∫ (sin2t)^2 dt=
=(625/4) ∫ (1- cos4t)dt/2=
=(625/8) ∫ (1-cos4t)dt=(625/8)*( t - (1/4) sin4t)+C=
(x/5=sint; t=arcsin(x/5))
=(625/8)*arcsin(x/5) - ( 625/32)*sin(4arcsin(x/5)) + C
4.
tg(x/2)=t
x/2=arctgt
x=2arctgt
dx=2dt/(1+t^2)
sinx=2t/(1+t^2)
cos=(1-t^2)/(1+t^2)
получим
∫ ((2dt)/(1+t^2))/((2-2t^2 -2t+5+5t^2)/(1+t^2))=
= ∫ 2dt/(3t^2-2t+7)= (2/3) ∫ dt/(t-(1/3))^2+(20/9))=
=(2/3)arctg(t-(1/3))/sqrt(20/9) + C=
=(2/3)arctg (3t-1)/2sqrt(5) + C, t=tg(x/2)