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

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

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}

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

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