If ∫ 1 x 1 - x 1 + x d x = g x + c , g 1 = 0 , then g 1 2 is equal to

Given,

$\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}dx=g\left(x\right)+c$

Put $x=\cos 2\theta$

$dx=-2\sin 2\theta ·d\theta$

$=\int \frac{1}{\cos 2\theta }\tan \theta \left(-4\sin \theta ·\cos \theta \right)d\theta$

$=\int \frac{1}{\cos 2\theta }\left(-4{\sin }^2\theta \right)d\theta$

$=-2\int \frac{1-\cos 2\theta }{\cos 2\theta }d\theta$

$=-\frac{2}{2}\ln \left|\sec 2\theta -\tan 2\theta \right|+2\theta +\mathrm{c}$

$=\ln \left|\sec 2\theta -\tan 2\theta \right|+2\theta +c$

$=\ln \left|\frac{1-\sin 2\theta }{\cos 2\theta }\right|+{\cos }^{-1}x+c$

$\underset{g\left(x\right)}{\underbrace{=\ln \left|\frac{1-\sqrt{1-x^2}}{x}\right|+{\cos }^{-1}x+c}}$

$\therefore g\left(1\right)=0\Rightarrow c=0$

So, $g\left(x\right)=\ln \left|\frac{1-\sqrt{1-x^2}}{x}\right|+{\cos }^{-1}x$

$g\left(\frac{1}{2}\right)=\ln \left|2-\sqrt{3}\right|+\frac{\pi }{3}$

$g\left(\frac{1}{2}\right)=\ln \left|\frac{\sqrt{3}-1}{\sqrt{3}+1}\right|+\frac{\pi }{3}$