Evaluate $\displaystyle\int x^2 \sqrt{a^6 - x^6} \; dx$
Let $I = \displaystyle\int x^2 \sqrt{a^6 - x^6} \; dx$ $\;\;\; \cdots$ (1)
Put $x^3 = u$ $\;\;\; \cdots$ (2a) $\implies$ $x^6 = u^2$ $\;\;\; \cdots$ (2b)
Differentiating equation (2a) gives
$3 x^2 dx = du$ $\implies$ $x^2 dx = \dfrac{du}{3}$ $\;\;\; \cdots$ (3)
$\therefore$ $\;$ In view of equations (2b) and (3), equation (1) becomes
$I = \dfrac{1}{3} \displaystyle\int \sqrt{a^6 - u^2} \; du$ $\;\;\; \cdots$ (4)
Put $u = a^3 \sin \theta$ $\;\;\; \cdots$ (5a) $\implies$ $u^3 = a^6 \sin^2 \theta$ $\;\;\; \cdots$ (5b)
Differentiating equation (5a) gives, $du = a^3 \cos \theta \; d\theta$ $\;\;\; \cdots$ (6)
$\therefore$ In view of equations (5b) and (6), equation (4) becomes
$\begin{aligned}
I & = \dfrac{1}{3} \int a^3 \cos \theta \sqrt{a^6 - a^6 \sin^2 \theta} \; d\theta \\\\
& = \dfrac{a^3}{3} \int \cos \theta \sqrt{a^6 \left(1 - \sin^2 \theta\right)} \; d\theta \\\\
& = \dfrac{a^6}{3} \int \cos^2 \theta \; d\theta \\\\
& = \dfrac{a^6}{3} \int \dfrac{1 + \cos 2 \theta}{2} \; d\theta \\\\
& = \dfrac{a^6}{6} \left(\int d \theta + \int \cos 2 \theta \; d \theta\right) \\\\
& = \dfrac{a^6}{6} \left(\theta + \dfrac{\sin 2 \theta}{2}\right) + c \\\\
& = \dfrac{a^6}{6} \left(\theta + \dfrac{2 \sin \theta \cos \theta}{2}\right) + c \\\\
& = \dfrac{a^6}{6} \left(\theta + \sin \theta \cos \theta\right) + c \;\;\; \cdots (7)
\end{aligned}$
Now, from equation (5a) we have
$\sin \theta = \dfrac{u}{a^3}$ $\;\;\; \cdots$ (8a) $\implies$ $\theta = \sin^{-1} \left(\dfrac{u}{a^3}\right)$ $\;\;\; \cdots$ (8b)
Also, $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \dfrac{u^2}{a^6}} = \dfrac{\sqrt{a^6 - u^2}}{a^3}$ $\;\;\; \cdots$ (8c)
$\therefore$ $\;$ In view of equations (8a), (8b) and (8c), equation (7) can be written as
$\begin{aligned}
I & = \dfrac{a^6}{6} \left\{\sin^{-1} \left(\dfrac{u}{a^3}\right) + \dfrac{u}{a^3} \times \dfrac{\sqrt{a^6 - u^2}}{a^3} \right\} + c \\\\
& = \dfrac{a^6}{6} \sin^{-1} \left(\dfrac{u}{a^3}\right) + \dfrac{u \sqrt{a^6 - u^2}}{6} + c \;\;\; \cdots (9)
\end{aligned}$
In view of equation (2a), equation (9) becomes
$I = \dfrac{a^6}{6} \sin^{-1} \left(\dfrac{x^3}{a^3}\right) + \dfrac{x^3 \sqrt{a^6 - x^6}}{6}$ + c