Question

Evaluate: $\int (\frac{1-\sqrt{x}}{1+\sqrt{x}}{)}^{1/2}\cdot \frac{dx}{x}$

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

Answer 1

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

$I=\int (\frac{1-\sqrt{x}}{1+\sqrt{x}}{)}^{1/2}\cdot \frac{dx}{x}$
Substituting $\sqrt{x}=\cos \theta$
$\frac{1}{2\sqrt{x}}dx=-\sin \theta$
$\Rightarrow dx=-2\sin \theta \cos \theta d\theta$
So, $I=-2\int ({\tan }^2\frac{\theta }{2}{)}^{1/2}\tan \theta d\theta$
$I=-2\int \tan \frac{\theta }{2}\tan \theta d\theta$
$I=-4\int \frac{{\sin }^2\left(\frac{\theta }{2}\right)}{\cos \theta }d\theta$
$=-2\int \frac{1-\cos \theta }{\cos \theta }d\theta$
$=-2\int (\sec \theta -1)d\theta$
$I=-2\log |\sec \theta +\tan \theta |+2\theta +C$
$I=-2\log |\frac{1+\sin \theta }{\cos \theta }|+2\theta +C$
$I=-2\log |\frac{1+\sqrt{1-x}}{\sqrt{x}}|+2{\cos }^{-1}\sqrt{x}+C$