Question

integrala din radical din x supra radical din 1-x^3

E x^3....cu x^2 știu și eu

Answer 1

Răspuns:

S.v. $t=x^3$, adica $x=\sqrt[3]{t}=t^{\frac{1}{3}}$, $dx=\frac{1}{3}t^{-\frac{2}{3}}dt$. Deci:

$\int \frac{\sqrt x}{\sqrt{1-x^3}}dx = \int \frac{t^{\frac{1}{6}}}{\sqrt{1-t}}\frac{1}{3}t^{-\frac{2}{3}}dt=\frac{1}{3}\int \frac{1}{\sqrt{1-t}}t^{\frac{1}{6}-\frac{2}{3}}dt =\frac{1}{3}\int \frac{1}{\sqrt{1-t}}t^{-\frac{1}{2}}dt=$

$=\frac{1}{3}\int \frac{1}{\sqrt{(1-t)t}}=\frac{1}{3}\int \frac{1}{\sqrt{t-t^2}}dt = \frac{1}{3}\int \frac{1}{\sqrt{\frac{1}{4}-(t-\frac{1}{2})^2}}dt=$

$=\frac{1}{3} arcsin(2(t-\frac{1}{2}))+\mathcal C = \frac{1}{3} arcsin(2t-1))+\mathcal C = \frac{1}{3} arcsin(2x^3-1))+\mathcal C.$