#### Explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.28 question 13

$\frac{x^{3}}{6} \sqrt{a^{6}-x^{6}}+\frac{a^{6}}{6} \sin ^{-1}\left(\frac{x^{3}}{a^{3}}\right)+c$

Hints:-

\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+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 \end{aligned}

Given:-

$\int x^{2} \sqrt{a^{6}-x^{6}} d x$

Solution:-

Let,

\begin{aligned} &I=\int x^{2} \sqrt{a^{6}-x^{6}} d x \\\\ &I=\int x^{2} \sqrt{a^{6}-\left(x^{3}\right)^{2}} d x \end{aligned}

Let, $x^{3}= t$

Differentiating both sides,

\begin{aligned} &\Rightarrow 3 x^{2} d x=d t \\\\ &\Rightarrow x^{2} d x=\frac{1}{3} d t \end{aligned}

Substituting  $x^{3}$  with t, we have

\begin{aligned} &I=\frac{1}{3} \int \sqrt{\left(a^{3}\right)^{2}-t^{2}} d t \\\\ &I=\frac{1}{2} \int \sqrt{\left(a^{3}\right)^{2}-t^{2}} d t \end{aligned}

As I match with the form

\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+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)\right\}+c \\\\ &I=\frac{t}{4} \sqrt{t^{2}+1}+\frac{1}{4} \log \left|t+\sqrt{t^{2}+1}\right|+c \end{aligned}

Putting the value of t i.e.t=x3

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