Evaluate the integral:


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. $t=x^{3}$

$\Rightarrow I=\frac{x^{3}}{6} \sqrt{a^{6}-x^{6}}+\frac{a^{6}}{6} \sin ^{-1}\left(\frac{x^{3}}{a^{3}}\right)+C$

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