Question

Solve:
${\int }_0^1\sqrt{\frac{1-x}{1+x}dx}$

Answer 1

We have,

$\int \sqrt{\frac{1-x}{1+x}}dx$

$=\int \sqrt{\frac{1-x}{1+x}}\frac{(1-x)}{(1-x)}dx$

$=\int \frac{1-x}{\sqrt{1+x^2}}dx$

$=\int \left[\frac{1}{\sqrt{1-x^2}}-\frac{x}{\sqrt{1-x^2}}\right]dx$

$={\sin }^{-1}x+\sqrt{1-x^2}+c$

Hence, this is the answer.