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$\int_0^1 \sqrt{\frac{1-x}{1+x}} d x$ is equal to
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2926 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{2}-1$
Let
$$
\begin{aligned}
I & =\int_0^1 \sqrt{\frac{1-x}{1+x}} d x \\
& =\int_0^1 \sqrt{\frac{1-x}{1+x}} \times \sqrt{\frac{1-x}{1-x}} d x \\
& =\int_0^1 \frac{1-x}{\sqrt{1-x^2}} d x \\
& =\int_0^1 \frac{1}{\sqrt{1-x^2}} d x-\int_0^1 \frac{x}{\sqrt{1-x^2}} d x
\end{aligned}
$$
$$
\begin{aligned}
& =\left[\sin ^{-1} x\right]_0^1+\left[\sqrt{1-x^2}\right]_0^1 \\
& =\left(\sin ^{-1} 1-\sin ^{-1} 0\right)+(\sqrt{1-1}-\sqrt{1-0}) \\
& =\frac{\pi}{2}-1
\end{aligned}
$$
$$
\begin{aligned}
I & =\int_0^1 \sqrt{\frac{1-x}{1+x}} d x \\
& =\int_0^1 \sqrt{\frac{1-x}{1+x}} \times \sqrt{\frac{1-x}{1-x}} d x \\
& =\int_0^1 \frac{1-x}{\sqrt{1-x^2}} d x \\
& =\int_0^1 \frac{1}{\sqrt{1-x^2}} d x-\int_0^1 \frac{x}{\sqrt{1-x^2}} d x
\end{aligned}
$$
$$
\begin{aligned}
& =\left[\sin ^{-1} x\right]_0^1+\left[\sqrt{1-x^2}\right]_0^1 \\
& =\left(\sin ^{-1} 1-\sin ^{-1} 0\right)+(\sqrt{1-1}-\sqrt{1-0}) \\
& =\frac{\pi}{2}-1
\end{aligned}
$$
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