Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18 point 28 Question 18 Maths Textbook Solution.

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

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:-

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

Solution:-

\begin{aligned} &\text { Let, } I=\int \sqrt{2 x-x^{2}} d x \\\\ &\therefore I=\int \sqrt{-\left(x^{2}-2(1) x\right)} d x \end{aligned}

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