Question

Evaluate: $\int {\left[\frac{1-\sqrt{x}}{1+\sqrt{x}}\right]}^{1/2}\frac{dx}{x}$

• A. $2{\cos }^{-1}\sqrt{x}-2\log \left[\frac{1+\sqrt{1-x}}{\sqrt{x}}\right]$
• B. $2{\sin }^{-1}\sqrt{x}+2\log \left[\frac{1+\sqrt{1-x}}{\sqrt{x}}\right]$
• C. $2{\cos }^{-1}\sqrt{x}+2\log \left[\frac{1-\sqrt{1+x}}{\sqrt{x}}\right]$
• D. $2{\sin }^{-1}\sqrt{x}-2\log \left[\frac{1-\sqrt{1+x}}{\sqrt{x}}\right]$

Answer 1

The correct option is A $2{\cos }^{-1}\sqrt{x}-2\log \left[\frac{1+\sqrt{1-x}}{\sqrt{x}}\right]$

Let $I=\int {\left[\frac{1-\sqrt{x}}{1+\sqrt{x}}\right]}^{1/2}\frac{dx}{x}$
Substitute $x={\cos }^{2}2\theta ,\frac{1-\cos 2\theta }{1+\cos 2\theta }={\tan }^2\theta \Rightarrow dx=-4\cos .\sin 2\theta d\theta$
$I=-4\int \frac{\sin \theta }{\cos \theta }.\frac{\cos 2\theta (2\sin \theta .\cos \theta )}{{\cos }^{2}2\theta }d\theta =-4\int \frac{2{\sin }^2\theta }{\cos 2\theta }d\theta$
$=-4\int \frac{(1-\cos 2\theta )}{\cos 2\theta }d\theta =-4\int (\sec 2\theta -1)d\theta =-4[\frac{1}{2}\log (\sec 2\theta +\tan 2\theta )-\theta ]$
$I=4\theta -2\log [\sec 2\theta +\sqrt{{\sec }^{2}2\theta -1}]$
Now put $\cos 2\theta =\sqrt{x}\Rightarrow \theta =\frac{1}{2}{\sec }^{-1}\frac{1}{\sqrt{x}}=\frac{1}{2}{\cos }^{-1}\sqrt{x}$
$I=2{\cos }^{-1}\sqrt{x}-2\log [\frac{1}{\sqrt{x}}+\sqrt{\frac{1}{x}-1}]=2{\cos }^{-1}\sqrt{x}-2\log \left[\frac{1+\sqrt{1-x}}{\sqrt{x}}\right]$