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

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

Answers (1)

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}

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