Evaluate $\int x^2 \sqrt{8 - x^6} \; dx$
Let $I = \int x^2 \sqrt{8 - x^6} \; dx$ $\;\;\; \cdots$ (1)
Let $x^3 = u$ $\;\;\; \cdots$ (2)
Differentiating equation (2) gives
$3 \; x^2 \; dx = du$ $\implies$ $x^2 \; dx = \dfrac{du}{3}$ $\;\;\; \cdots$ (2a)
Also, $x^6 = \left(x^3\right)^2 = u^2$ $\;\;\; \cdots$ (2b)
$\therefore$ In view of equations (2a) and (2b), equation (1) becomes
$\begin{aligned}
I & = \dfrac{1}{3} \int \sqrt{8 - u^2} \; du \\\\
& = \dfrac{1}{3} \int \sqrt{\left(2 \sqrt{2}\right)^2 - u^2} \; du \\\\
& \left[\text{Note: } \int \sqrt{a^2 - x^2} \; dx = \dfrac{x}{2} \; \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left(\dfrac{x}{a}\right) + c \right] \\\\
& = \dfrac{1}{3} \left[\dfrac{u}{2} \; \sqrt{\left(2 \sqrt{2}\right)^2 - u^2} + \dfrac{\left(2 \sqrt{2}\right)^2}{2} \sin^{-1} \left(\dfrac{u}{2 \sqrt{2}}\right)\right] + c \\\\
& = \dfrac{1}{3} \left[\dfrac{x^3}{2} \; \sqrt{8 - x^6} + 4 \sin^{-1} \left(\dfrac{x^3}{2 \sqrt{2}}\right)\right] + c \;\;\; \left[\text{From equation (2)}\right]
\end{aligned}$