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

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

Answers (1)

Answer:-

\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}

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