#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 89 Maths Textbook Solution.

$\inline \dpi{100} \dpi{100} -\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)\right]+c$

Hint:

You have to find value of A & B.

Given:

$\dpi{100} \int x \sqrt{1+x-x^{2}} d x$

Solution:

$\dpi{100} \text { let } x=A\left(\left(\frac{d}{d x}\right)\left(1+x-x^{2}\right)\right)+B$

$\dpi{100} x=A(1-2 x)+B$

$\dpi{100} x=A+B-2 A x$

A + B = 0 ,    -2A = 1

$\dpi{100} B=\frac{1}{2} \quad A=-\frac{1}{2}$

$\dpi{100} \mathrm{I}=\int\left[-\frac{1}{2}(1-2 x)+\frac{1}{2}\right) \sqrt{1+x-x^{2}} d x$

$\dpi{100} =-\frac{1}{2} \int(1-2 x) \sqrt{1+x-x^{2}} d x+\frac{1}{2} \sqrt{1+x-x^{2}} d x$

$\dpi{100} I=-\frac{1}{2} \int \sqrt{t} d t \ldots .\left(\sqrt{1+x-x^{2}}=\mathrm{t}\right)$

$\dpi{100} =-\frac{1}{3} t^{\frac{3}{2}}=-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+c$

$\dpi{100} I I=\frac{1}{2} \int \sqrt{1+x-x^{2}} d x$

$\dpi{100} =\frac{1}{2} \int \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}} d x$                                                    $\dpi{100} \left[\because x-\frac{1}{2}=z, d x=d z\right]$

$\dpi{100} =\frac{1}{2} \int \sqrt{\left(\frac{\sqrt{5}}{2}\right)^{2}-(z)^{2}} d z$

$\dpi{100} =\frac{1}{2}\left[\frac{1}{2} z \sqrt{\frac{\sqrt{5}}{4}-z^{2}}+\frac{1}{2} \cdot \frac{\sqrt{5}}{4} \sin ^{-1}\left(\frac{2 z}{\sqrt{5}}\right)+c\right]$

$\dpi{100} =\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)+c\right]$

$\dpi{100} \int x \sqrt{1+x-x^{2}}=I+I I$

$\dpi{100} I=-\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+c+\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)+c\right]$

$\inline \dpi{100} I=-\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)\right]+C$