$\int_0^1 \sqrt{\frac{1-x}{1+x}} d x=$

Question

Aicharya · Accepted Answer

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

$\int_0^1 \sqrt{\frac{1-x}{1+x}} d x=$

Solution