#### Provide solution for RD Sharma maths class12 Chapter Definite Integrals exercise 19.2 question 47.

Answer :   $\frac{19}{99}$

Hint: Use in definite formula and the given limits to solve this integral

Given:

$\int_{4}^{9}\frac{\sqrt{x}}{[30-x^{\frac{3}{2}}]^2}dx$

Solution:

$\int_{4}^{9}\frac{\sqrt{x}}{[30-x^{\frac{3}{2}}]^2}dx$

Put $(30-x^{\frac{3}{2}})=t$

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

when x=4  then t=22

when x=9  then t=3  then

\begin{aligned} &\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{3 / 2}\right)^{2}} d x \\ &=\int_{22}^{3} \frac{\sqrt{x}}{t^{2}} \cdot \frac{-2 d t}{3 \sqrt{x}} \\ &=\frac{-2}{3} \int_{22}^{3} t^{2} d t \\ &=-\frac{2}{3}\left[\frac{t^{-2+1}}{-2+1}\right]_{22}^{3} \end{aligned}

\begin{aligned} &=\frac{-2}{3}\left[\frac{t^{-1}}{-1}\right]_{22}^{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=\frac{2}{3}\left[\frac{1}{t}\right]_{22}^{3} \\ &=\frac{2}{3}\left[\frac{1}{3}-\frac{1}{22}\right] \\ &=\frac{2}{3}\left[\frac{22-3}{66}\right] \\ &=\frac{2}{3} \cdot \frac{19}{66} \\ &=\frac{19}{99} \end{aligned}