#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 105 Maths Textbook Solution.

Answer:          $\sqrt{1-x^{2}}\left(\frac{x}{2}-1\right)-\frac{1}{2} \sin ^{-1}(x)+C$

Hint: to solve this equation, we have to assume u as $\cos \Theta$

Given: $\int x \sqrt{\frac{1-x}{1+x}} d x$

Solution:

$I=\int x \sqrt{\frac{1-x}{1+x}} d x$

$I=\int x \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} d x$

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

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

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

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

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

$I=\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1}(x)+C_{1}-\sqrt{1-x^{2}}+C_{2}-\sin ^{-1}(x)$

$+C_{3}\left[\because \int \frac{x}{\sqrt{1-x^{2}}} d x=-\sqrt{1-x^{2}}+C_{2}\right]$

$I=\sqrt{1-x^{2}}\left(\frac{x}{2}-1\right)-\frac{1}{2} \sin ^{-1}(x)+C$