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  • Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.28 Question 11

Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.28 Question 11

Answers (1)

Answer:-

The answer is
\frac{1}{4}(2 x+1) \sqrt{3-2 x-2 x^{2}}+\frac{7 \sqrt{2}}{8} \sin ^{-1} \frac{2 x+1}{\sqrt{7}}+c

Hints:-

\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+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{3-2 x-2 x^{2}} d x

Solution:-

Let,

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

We have

I=\int \sqrt{\frac{7}{4}-2\left(x+\frac{1}{2}\right)^{2}} d x=\int \sqrt{2} \sqrt{\left(\frac{\sqrt{7}}{2}\right)^{2}-\left(x+\frac{1}{2}\right)^{2}} d x

As I match with the form

\begin{aligned} &\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c \\\\ &I=\sqrt{2}\left\{\frac{x+\frac{1}{2}}{2} \sqrt{\left(\frac{\sqrt{7}}{2}\right)^{2}-\left(x+\frac{1}{2}\right)^{2}}\right\}+\frac{\frac{7}{4}}{2} \sin ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{7}}{2}}\right)+c \end{aligned}

I=\frac{1}{4}(2 x+1) \sqrt{2\left\{\left(\frac{\sqrt{7}}{2}\right)^{2}-\left(x+\frac{1}{2}\right)^{2}\right\}}+\frac{7 \sqrt{2}}{8} \sin ^{-1}\left(\frac{2 x+1}{\sqrt{7}}\right)+c

I=\frac{1}{4}(2 x+1) \sqrt{3-2 x-2 x^{2}}+\frac{7 \sqrt{2}}{8} \sin ^{-1}\left(\frac{2 x+1}{\sqrt{7}}\right)+c

Posted by

infoexpert27

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