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

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

Answers (1)

best_answer

Answer:-

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}

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