Question

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

Answer 1

2 \ln\left(\frac{1-\sqrt{1-x}}{\sqrt{x}}\right) - 2 \arcsin \sqrt{x} + C

Answer 2

Let $x = \sin^2 \theta$. Then $dx = 2 \sin \theta \cos \theta d\theta$. $\sqrt{x} = \sin \theta$. $\frac{dx}{x} = \frac{2 \sin \theta \cos \theta d\theta}{\sin^2 \theta} = 2 \cot \theta d\theta$. The integrand becomes: $\left(\frac{1-\sin \theta}{1+\sin \theta}\right)^{1/2} \cdot 2 \cot \theta$ $\left(\frac{(1-\sin \theta)^2}{\cos^2 \theta}\right)^{1/2} \cdot 2 \frac{\cos \theta}{\sin \theta}$ For $\theta \in (0, \pi/2]$, this is: $\frac{1-\sin \theta}{\cos \theta} \cdot 2 \frac{\cos \theta}{\sin \theta} = \frac{2(1-\sin \theta)}{\sin \theta} = 2(\csc \theta - 1)$ The integral is: $\int 2(\csc \theta - 1) d\theta = 2(\ln|\csc \theta - \cot \theta| - \theta) + C$ Substitute back $\theta = \arcsin \sqrt{x}$: $\csc \theta = \frac{1}{\sqrt{x}}$ $\cot \theta = \frac{\sqrt{1-x}}{\sqrt{x}}$ $2\left(\ln\left|\frac{1}{\sqrt{x}} - \frac{\sqrt{1-x}}{\sqrt{x}}\right| - \arcsin \sqrt{x}\right) + C$ $2\left(\ln\left|\frac{1-\sqrt{1-x}}{\sqrt{x}}\right| - \arcsin \sqrt{x}\right) + C$ $2 \ln\left(\frac{1-\sqrt{1-x}}{\sqrt{x}}\right) - 2 \arcsin \sqrt{x} + C$