Evaluate $\int \left( \frac{1-\sqrt{x}}{1+\sqrt{x}} \right)^{1/2} \frac{dx}{x}$.

Question

Professor PC Thomas Classes · Answer

Let $x = \cos^2 	heta$, so $dx = -2\cos 	heta \sin 	heta d	heta$. The integral becomes $\int 	an(	heta/2) \frac{-2\cos	heta\sin	heta}{\cos^2	heta} d	heta = -4\int \frac{\sin^2(	heta/2)}{\cos	heta} d	heta = -2 \int \frac{1-\cos	heta}{\cos	heta} d	heta = -2\int (\sec	heta - 1)d	heta$. This evaluates to $-2\ln|\sec	heta + 	an	heta| + 2	heta + C$. Substitute back $	heta = \cos^{-1}\sqrt{x}$, $\sec	heta = \frac{1}{\sqrt{x}}$, and $	an	heta = \frac{\sqrt{1-x}}{\sqrt{x}}$ to get Option A.