The value of the integral $\int_0^1 \sqrt{\frac{1-x}{1+x}} d x$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \int_0^1 \sqrt{\frac{1-x}{1+x}} d x=\int_0^1 \frac{1-x}{\sqrt{1-x^2}} d x=\int_0^1 \frac{d x}{\sqrt{1-x^2}}+\int_0^1 \frac{-2 x}{2 \sqrt{1-x^2}} d x \ & =\left[\sin ^{-1} xight]_0^1+\left[\sqrt{1-x^2}ight]_0^1 \ & =\left\{\sin ^{-1}(1)-\sin ^{-1}(0)ight\}+\left\{\sqrt{1-1^2}-\sqrt{1-0^2}ight\} \ & =\frac{\pi}{2}-0+0-1 \ & =\frac{\pi}{2}-1\end{aligned}$

The value of the integral $\int_0^1 \sqrt{\frac{1-x}{1+x}} d x$ is

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