#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.28 Question 4

$\frac{1}{8}(4 x-1) \sqrt{1+x-2 x^{2}}+\frac{9 \sqrt{2}}{32} \sin ^{-1}(2 x-1)+c$

Multiplying by $\sqrt{2}$.

Given:-

$\int \sqrt{1+x-2 x^{2}} d x$

Solution:-

$\int \sqrt{1+x-2 x^{2}} d x$

By multiplying by $\sqrt{2}$

\begin{aligned} &=\sqrt{2} \int \sqrt{\frac{1}{2}+\frac{x}{2}-x^{2}} d x \\\\ &=\sqrt{2} \int \sqrt{\frac{9}{16}-\left(\frac{1}{16}-\frac{x}{2}+x^{2}\right)} d x \\\\ &=\sqrt{2} \int \sqrt{\left(\frac{3}{4}\right)^{2}-\left(x-\frac{1}{4}\right)^{2}} d x \end{aligned}

\begin{aligned} &=\sqrt{2}\left\{\left(\frac{\left(x-\frac{1}{4}\right)}{2}\right) \sqrt{\frac{1}{2}+\frac{x}{2}-x^{2}}+\frac{9}{32} \sin ^{-1}\left(\frac{x-\frac{1}{4}}{3 / 4}\right)^{2}\right\}+c \\\\ &=\frac{1}{8}(4 x-1) \sqrt{1+x-2 x^{2}}+\frac{9 \sqrt{2}}{32} \sin ^{-1}\left(\frac{4 x-1}{3}\right)+c \end{aligned}

Using the formula

$\int \sqrt{a^{2}-x^{2}}=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+c$