Question

The value of integral $\underset{0}{\overset{1}{\int }}\sqrt{\frac{1-x}{1+x}}dx$ is

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

Answer 1

Answer: (B)

Let $I=\underset{0}{\overset{1}{\int }}\sqrt{\frac{1-x}{1+x}}dx$
$=\underset{0}{\overset{1}{\int }}\frac{1-x}{\sqrt{1-x^2}}dx$
$=\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{1-x^2}}dx-\underset{0}{\overset{1}{\int }}\frac{x}{\sqrt{1+x^2}}dx$
$={\left[{\sin }^{-1}x\right]}_0^1-\underset{0}{\overset{1}{\int }}\frac{x}{\sqrt{1-x^2}}dx$
Put $t^2=1-x^2$
$\Rightarrow 2tdt=-2xdx$
$\Rightarrow tdt=-xdx$
$\therefore I=\left({\sin }^{-1}1-{\sin }^{-1}0\right)+\underset{1}{\overset{0}{\int }}\frac{t}{t}dt$
$=\frac{\pi }{2}+{\left[t\right]}_1^0=\frac{\pi }{2}-1$
Let $I=\underset{0}{\overset{1}{\int }}\sqrt{\frac{1-x}{1+x}}dx$
Put $x=\cos 2\theta$
$\Rightarrow dx=-2\sin 2\theta d\theta$
$\therefore I=-\underset{\pi /4}{\overset{0}{\int }}\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}\cdot 2\sin \theta d\theta$
$=-2\underset{\pi /4}{\overset{0}{\int }}\frac{\sin \theta }{\cos \theta }\cdot 2\sin \theta \cos \theta d\theta$
$=-2\underset{\pi /4}{\overset{0}{\int }}(1-\cos 2\theta )d\theta$
$=-2{\left[\theta -\frac{\sin 2\theta }{2}\right]}_{\pi /4}^0$
$=-2\left[0-\left(\frac{\pi }{4}-\frac{1}{2}\right)\right]$
$=\frac{\pi }{2}-1$