Răspuns:
Facem schimbarea de variabila: [tex]y=x^2[/tex].
[tex]dy = (x^2)'dx=2xdx[/tex].
Daca [tex]x=-\frac{1}{\sqrt 2}[/tex], atunci [tex]y=\frac{1}{2}[/tex]
Daca x=0 atunci y =0. Rezulta ca:
[tex] \int_{-\frac{1}{\sqrt 2}}^0\frac{2x}{\sqrt{1-x^4}} dx = \int_{\frac{1}{2}}^0 \frac{dy}{\sqrt{1-y^2}} = arcsin(y)|_{1/2}^0 = arcsin(0)-arcsin(1/2) =-\frac{\pi}{6}. [/tex]
[tex]arcsin(1/2)=\frac{\pi}{6}[/tex] pentru ca [tex]sin(\pi/6)=1/2[/tex].
arcsin(0)=0 pentru ca sin(0)=0.
