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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


\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}



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