#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.21 Question 14 Maths Textbook Solution.

Answer: $\sin ^{-1} x+\sqrt{1-x^{2}}+c$

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

Hint: Simplify the given function

Solution:

\begin{aligned} I &=\int \sqrt{\frac{1-x}{1+x}} d x \\ I &=\int \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x \text { (rationalizing ta e denominator) } \\ &=\int \sqrt{\frac{(1-x)^{2}}{1-x^{2}}} d x \quad\left[(a+b)(a-b)=\left(a^{2}-b^{2}\right)\right] \\ I &=\int \frac{1-x}{\sqrt{1-x^{2}}} d x \end{aligned}

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

\begin{aligned} &{\left[\begin{array}{c} \int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1}\left(\frac{x}{a}\right)+c \\ \int(f(x))^{n} f^{1}(x) d x=\frac{[f(x)]^{n+1}}{n+1}+c \end{array}\right]} \\ &I=\sin ^{-1} x+\sqrt{1-x^{2}}+c \end{aligned}