Задача №1324
Условие
Найти интеграл \(\int\frac{x^3dx}{\sqrt{1-x^8}}\).
Решение
\[
\int\frac{x^3dx}{\sqrt{1-x^8}}
=\int\frac{x^3dx}{\sqrt{1-\left(x^4\right)^2}}
=\left[\begin{aligned}& d\left(x^4\right)=4x^3dx;\\& x^3dx=\frac{1}{4}d\left(x^4\right).\end{aligned}\right]
=\frac{1}{4}\int\frac{d\left(x^4\right)}{\sqrt{1-\left(x^4\right)^2}}
=\frac{1}{4}\arcsin{x^4}+C
\]
Ответ:
\(\frac{1}{4}\arcsin{x^4}+C\)