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  • Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 17

Provide Solution for RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 17

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Answer:  \frac{\pi}{2}-1

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

Hint: Do rationalization and then integrate by parts

Solution:

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

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

=\int_{0}^{1} \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x=\int_{0}^{1} \sqrt{\frac{(1-x)^{2}}{1-x^{2}}} d x

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

=\left(\sin ^{-1} x\right)_{0}^{1}-\left(-\frac{1}{2} \sqrt{\frac{1-x^{2}}{\frac{1}{2}}}\right)_{0}^{1}

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

=\left(\frac{\pi}{2}-1\right)

 

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