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  • Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 10

Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 10

Answers (1)

Answer:

\frac{\pi}{2}-1

Given:

\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x

Hint:

Explanation:  

Let

\begin{aligned} &I=\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x \\\\ &=\int_{0}^{1} \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x \end{aligned}

\begin{aligned} &=\int_{0}^{1} \sqrt{\frac{(1-x)^{2}}{1-x^{2}}} d x \\\\ &=\int_{0}^{1} \frac{1-x}{\sqrt{1-x^{2}}} d x \end{aligned}

\begin{aligned} &=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x-\int_{0}^{1} x\left(1-x^{2}\right)^{-1 / 2} d x \\\\ &=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x+\int_{0}^{1}\left(1-x^{2}\right)^{-1 / 2}(-2 x) d x \end{aligned}

\begin{aligned} &=\left[\sin ^{-1} x\right]_{0}^{1}+\frac{1}{2}\left[\frac{\left(1-x^{2}\right)^{1 / 2}}{1 / 2}\right]_{0}^{1} \\\\ &=\left[\sin ^{-1}(1)-\sin ^{-1}(0)\right]+\frac{1}{2}\left[0-\frac{1}{1 / 2}\right] \end{aligned}

\begin{aligned} &=\left(\frac{\pi}{2}-0\right)+\frac{1}{2}(-2) \\\\ &=\frac{\pi}{2}-1 \end{aligned}

 

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