If int frac{1}{x} sqrt{frac{1-x}{1+x}} dx = g | vedclass

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(A) Let $I = \int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} dx$. Put $x = \cos 2	heta$,so $dx = -2 \sin 2	heta d	heta$.Then $I = \int \frac{1}{\cos 2	het

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$\log_{e}\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right) + \frac{\pi}{3}$

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(A) Let $I = \int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} dx$. Put $x = \cos 2	heta$,so $dx = -2 \sin 2	heta d	heta$.Then $I = \int \frac{1}{\cos 2	heta} \sqrt{\frac{1-\cos 2	heta}{1+\cos 2	heta}} (-2 \sin 2	heta) d	heta = \int \frac{1}{\cos 2	heta} 	an 	heta (-4 \sin 	heta \cos 	heta) d	heta$.$I = \int \frac{-4 \sin^2 	heta}{\cos 2	heta} d	heta = -2 \int \frac{1-\cos 2	heta}{\cos 2	heta} d	heta = -2 \int (\sec 2	heta - 1) d	heta$.$I = -2 \left( \frac{1}{2} \ln |\sec 2	heta + 	an 2	heta| - 	heta ight) + c = -\ln |\sec 2	heta + 	an 2	heta| + 2	heta + c$.Since $\cos 2	heta = x$,$	an 2	heta = \frac{\sqrt{1-x^2}}{x}$ and $\sec 2	heta = \frac{1}{x}$,we have $g(x) = -\ln \left| \frac{1+\sqrt{1-x^2}}{x} ight| + \cos^{-1} x + c$.Using $g(1) = 0$,we find $c = 0$. Thus $g(x) = \ln \left| \frac{x}{1+\sqrt{1-x^2}} ight| + \cos^{-1} x = \ln \left| \frac{x(1-\sqrt{1-x^2})}{x^2} ight| + \cos^{-1} x = \ln \left| \frac{1-\sqrt{1-x^2}}{x} ight| + \cos^{-1} x$.For $x = 1/2$,$g(1/2) = \ln |2 - \sqrt{3}| + \cos^{-1}(1/2) = \ln \left( \frac{\sqrt{3}-1}{\sqrt{3}+1} ight) + \frac{\pi}{3}$.

If $\int \frac{1}{x} \sqrt{\frac{1-x}{1+x}} dx = g(x) + c$ and $g(1) = 0$,then $g\left(\frac{1}{2}\right)$ is equal to

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