Evaluate $\int\limits_{0}^{1/2} \frac{dx}{(1 + x^2)\sqrt{1 - x^2}}$

$\frac{1}{\sqrt{2}} \tan^{-1} \left( \sqrt{\frac{2}{3}} \right)$

The correct answer is Option (1) → $\frac{1}{\sqrt{2}} \tan^{-1} \left( \sqrt{\frac{2}{3}} \right)$

Let $I = \int\limits_{0}^{1/2} \frac{dx}{(1 + x^2)\sqrt{1 - x^2}}$

Put $x = \sin \theta \Rightarrow dx = \cos \theta d\theta$

As $x \to 0, \text{ then } \theta \to 0$ and $x \to \frac{1}{2}, \text{ then } \theta \to \frac{\pi}{6}$

$∴I = \int\limits_{0}^{\pi/6} \frac{\cos \theta}{(1 + \sin^2 \theta)\cos \theta} d\theta = \int\limits_{0}^{\pi/6} \frac{1}{1 + \sin^2 \theta} d\theta  [∵1 - \sin^2 \theta = \cos^2 \theta]$

$= \int\limits_{0}^{\pi/6} \frac{1}{\cos^2 \theta (\sec^2 \theta + \tan^2 \theta)} d\theta$

$= \int\limits_{0}^{\pi/6} \frac{\sec^2 \theta}{\sec^2 \theta + \tan^2 \theta} d\theta$

$= \int\limits_{0}^{\pi/6} \frac{\sec^2 \theta}{1 + \tan^2 \theta + \tan^2 \theta} d\theta  [∵\sec^2 \theta = 1 + \tan^2 \theta]$

$= \int\limits_{0}^{\pi/6} \frac{\sec^2 \theta}{1 + 2\tan^2 \theta} d\theta$

Again, put $\tan \theta = t \Rightarrow \sec^2 \theta d\theta = dt$

As $\theta \to 0, \text{ then } t \to 0$ and $\theta \to \frac{\pi}{6}, \text{ then } t \to \frac{1}{\sqrt{3}}$

$∴I = \int\limits_{0}^{1/\sqrt{3}} \frac{dt}{1 + 2t^2} = \frac{1}{2} \int\limits_{0}^{1/\sqrt{3}} \frac{dt}{\left( \frac{1}{\sqrt{2}} \right)^2 + t^2}$

$= \frac{1}{2} \left[ \frac{1}{1/\sqrt{2}} \tan^{-1} \frac{t}{1/\sqrt{2}} \right]_0^{1/\sqrt{3}} = \frac{1}{\sqrt{2}} [\tan^{-1}(\sqrt{2}t)]_0^{1/\sqrt{3}}$

$= \frac{1}{\sqrt{2}} \left[ \tan^{-1} \sqrt{\frac{2}{3}} - 0 \right] = \frac{1}{\sqrt{2}} \tan^{-1} \left( \sqrt{\frac{2}{3}} \right)$