Evaluate the integral:
$\int x^{2} \sqrt{a^{6}-x^{6}} d x$
Key points to solve the problem:
- Such problems require the use of method of substitution along with method of integration by parts. By
method of integration by parts if we have $\int \mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{d} \mathrm{x}=\mathrm{f}(\mathrm{x}) \int \mathrm{g}(\mathrm{x}) \mathrm{dx}-\int \mathrm{f}^{\prime}(\mathrm{x})\left(\int \mathrm{g}(\mathrm{x}) \mathrm{dx}\right) \mathrm{dx}$
- To solve the integrals of the form: $\int \sqrt{a x^{2}+b x+c} d x$ after applying substitution and integration by parts we have direct formulae as described below:
$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$
$\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C$
$\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+C$
Let, $I=\int x^{2} \sqrt{a^{6}-x^{6}} d x=\int x^{2} \sqrt{a^{6}-\left(x^{3}\right)^{2}} d x$
Let, $x^{3}=t$
Differentiating both sides:
$\Rightarrow 3 x^{2} d x=d t$
$\Rightarrow x^{2} d x=1 / 3 d t$
Substituting $x^{3}$ with $t$, we have:
$\therefore I=\frac{1}{3} \int \sqrt{\left(a^{3}\right)^{2}-t^{2}} d t=\int \sqrt{\left(a^{3}\right)^{2}-t^{2}} d t$
As I match with the form: $\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$
$\therefore I=\frac{1}{3}\left\{\frac{t}{2} \sqrt{a^{6}-(t)^{2}}+\frac{a^{6}}{2} \sin ^{-1}\left(\frac{t}{a^{3}}\right)+C\right\}$
Putting the value of $t$ i.e. $t=x^{3}$
$\Rightarrow I=\frac{x^{3}}{6} \sqrt{a^{6}-x^{6}}+\frac{a^{6}}{6} \sin ^{-1}\left(\frac{x^{3}}{a^{3}}\right)+C$