#### Explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.28 question 15

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

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:-

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

Solution:-

Let,

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

We have

$I=\int \sqrt{(a)^{2}-(x-a)^{2}} d x$

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