#### Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 69 maths textbook solution

Answer: $-\frac{20}{3}\left(5-x^{2}\right)^{\frac{3}{2}}+\frac{4}{5}\left(5-x^{2}\right)^{\frac{5}{2}}+c$

Hint: Use substitution method to solve this integral

Given: $\int 4 x^{3} \sqrt{5-x^{2}} \; d x$

Solution:

\begin{aligned} &\text { Let } I=\int 4 x^{3} \sqrt{5-x^{2}} \; d x \\ &\text { Put } 5-x^{2}=t^{2} \Rightarrow-2 x \; d x=2 t\; d t \\ &\Rightarrow d x=\frac{t}{-x} d t \text { then } \end{aligned}

$I=\int 4 x^{3} \sqrt{t^{2}} \frac{t \cdot d t}{-x}=-\int 4 x^{2} \cdot t . t \; d t$

$=-4 \int\left(5-t^{2}\right) t^{2} d t=-4 \int\left(5 t^{2}-t^{2} \cdot t^{2}\right) d t \quad\left[\because 5-x^{2}=t^{2} \Rightarrow x^{2}=5-t^{2}\right]$

$=-4 \int\left(5 t^{2}-t^{4}\right) d t$

$=-20 \int t^{2} d t+4 \int t^{4} d t$

$=-20 \cdot \frac{t^{2+1}}{2+1}+4 \cdot \frac{t^{4+1}}{4+1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$

$=-20 \cdot \frac{t^{3}}{3}+4 \frac{t^{5}}{5}+c$

$=-\frac{20}{3}\left(\sqrt{5-x^{2}}\right)^{3}+\frac{4}{5}\left(\sqrt{5-x^{2}}\right)^{5}+c \quad\left[\because t^{2}=5-x^{2} \Rightarrow t=\sqrt{5-x^{2}}\right]$

$=-\frac{20}{3}\left(5-x^{2}\right)^{\frac{3}{2}}+\frac{4}{5}\left(5-x^{2}\right)^{\frac{5}{2}}+c$