Question

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

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

Answer 1

Answer: (C)

$I={\int }_0^1[\sqrt{\frac{1-x}{1+x}}\times \frac{\sqrt{1-x}}{\sqrt{1-x}}]dx(rationalisingthedenominator)$
$$\begin{gathered} ={\int }_0^1\frac{1-x}{\sqrt{1-x^2}}dx \\ ={\int }_0^1\frac{1}{\sqrt{1-x^2}}dx-{\int }_0^1\frac{xdx}{\sqrt{1-x^2}} \end{gathered}$$
$\Rightarrow I=[si{n}^{-1}x{]}_0^1-\frac{1}{2}{\int }_0^1\frac{2x}{\sqrt{1-x^2}}dx$
$=[\frac{\pi }{2}-0]+\frac{1}{2}[2\sqrt{1-x^2}{]}_0^1$
$$\begin{gathered} =\frac{\pi }{2}+[0-1] \\ =\frac{\pi }{2}-1 \end{gathered}$$