Задача №1558
Условие
Найти интеграл \(\int\frac{\sqrt{1-x^3}}{x^2\sqrt{x}}dx\).
Решение
\[
\int\frac{\sqrt{1-x^3}}{x^2\sqrt{x}}dx
=\left[\begin{aligned}
& z=\sqrt{x^3};\;dx=\frac{2dz}{3\sqrt[3]{z}}.\\
& 0\lt{z}\le{1}.
\end{aligned}\right]
=\frac{2}{3}\int\frac{\sqrt{1-z^2}}{z^2}dz
=\left[z=\sin{t};\;0\lt{t}\le\frac{\pi}{2}.\right]=\\
=\frac{2}{3}\int\frac{\cos^2{t}}{\sin^2{t}}dt
=\frac{2}{3}\int\left(\frac{1}{\sin^2{t}}-1\right)dt
=-\frac{2}{3}\ctg{t}-\frac{2}{3}t+C
=-\frac{2}{3}\sqrt{\frac{1-x^3}{x^3}}-\frac{2}{3}\arcsin\sqrt{x^3}+C.
\]
Ответ:
\(-\frac{2}{3}\sqrt{\frac{1-x^3}{x^3}}-\frac{2}{3}\arcsin\sqrt{x^3}+C\)