Задача 26652 ...
Условие
в) Интеграл 1/ x^2√x^2+9 dx
Все решения
dx=2costdt
= ∫ sqrt(4-4sin^2t) * 2 costdt=4 ∫ cos^2tdt=4 ∫ (1+cos2t)dt/2=
=2 ∫ dt+2 ∫ cos2tdt=2t+(1/2)*(sin2t)+C
sint=x/2 ⇒ t=arcsin(x/2)
cost=sqrt(1-sin^2t)=sqrt(1-(x/2)^2)=sqrt(4-x^2)/2
sin2t=2sint*cost=2*(x/2)*(sqrt(4-x^2)/2)=(1/2)x*sqrt(4-x^2)
О т в е т. 2*arcsin(x/2) +(1/4)*x*sqrt(4-x^2) + C
x=3tgt ⇒ sqrt(x^2+9)=sqrt(9tg^2t+9)=3sqrt(tg^2t+1)=(3/cost)
dx=3dt/cos^2t
= ∫ (1/9tg^2t)*(cost/3)*(3dt/cos^2t)=
=(1/9)∫dt/(tg^2t*cost)=(1/9)∫costdt/sin^2t=
=(1/9)∫d(sint)/sin^2t=(1/9)∫ sin^(-2)td(sint)=
=(1/9)*(-1/sint) + C
tgt=(x/3)
1+tg^2t=1+(x/3)^2
1/cos^2t=(x^2+9)/9
cost=3/sqrt(x^2+9)
sin^2t=1-cos^2t=1-(9/(x^2+9))=x^2/(x^2+9)
sint=x/sqrt(x^2+9)
О т в е т. (-1/9)*(sqrt(x^2+9)/x) + C