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

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

Answers (1)

Answer:-

\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

Posted by

infoexpert27

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