Question

The value of integral ${\int }_0^1\sqrt{\frac{1-x}{1+x}}dx$ is

• A. $\frac{\pi }{2}+1$
• B. $\frac{\pi }{2}-1$
• C. $-1$
• D. $1$

Answer 1

Answer: (A)

$\frac{\pi }{2}+1$

Let $I={\int }_0^1\sqrt{\frac{1-x}{1+x}}dx$

$={\int }_0^1\frac{1-x}{\sqrt{1-x^2}}dx$

$={\int }_0^1\frac{1}{\sqrt{1-x^2}}dx-{\int }_0^1\frac{x}{\sqrt{1-x^2}}dx$

$=[si{n}^{-1}x{]}_0^1-{\int }_0^1\frac{x}{\sqrt{1-x^2}}dx$

Put $t^2=1-x^2\Rightarrow 2tdt=-2xdx$
$\Rightarrow tdt=-xdx$
$\therefore I=(si{n}^{-1}1-si{n}^{-1}0)+{\int }_0^1\frac{t}{t}dt$

$=\frac{\pi }{2}+[t{]}_0^1=\frac{\pi }{2}+1$

Hence, option A is correct.