#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.28 Question 16 Maths Textbook Solution.

$I=\frac{x}{2} \sqrt{3-x^{2}}+\frac{3}{2} \sin ^{-1}\left(\frac{x}{\sqrt{3}}\right)+c$

Hint:-

Using the formula

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

Given:-

$\int \sqrt{3-x^{2}} d x$

Solution:-

\begin{aligned} &\text { Let, } I=\int \sqrt{3-x^{2}} d x \\ &\therefore I=\int \sqrt{3-x^{2}} d x=\int \sqrt{(\sqrt{3})^{2}-x^{2}} d x \end{aligned}

As I match with the form

\begin{aligned} &\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+c \\\\ &I=\frac{x}{2} \sqrt{3-x^{2}}+\frac{3}{2} \sin ^{-1}\left(\frac{x}{\sqrt{3}}\right)+c \end{aligned}