Evaluate : \(\int_0^{\frac{1}{2}} \frac{dx}{(1 - 2x^2)\cdot\sqrt{1 - x^2}}\)

Answer: \(\frac{1}{2}\log(2 + \sqrt{3})\)

Put \(x = \sin\theta\), so \(dx = \cos\theta\,d\theta\) and \(\sqrt{1 - x^2} = \cos\theta\). When \(x = 0,\ \theta = 0\); when \(x = \frac{1}{2},\ \theta = \frac{\pi}{6}\).

\(I = \int_0^{\pi/6} \frac{\cos\theta\,d\theta}{(1 - 2\sin^2\theta)\cos\theta} = \int_0^{\pi/6} \frac{d\theta}{\cos 2\theta} = \int_0^{\pi/6} \sec 2\theta\,d\theta\)

\(= \frac{1}{2}\left[\log|\sec 2\theta + \tan 2\theta|\right]_0^{\pi/6} = \frac{1}{2}\left[\log\left(\sec\frac{\pi}{3} + \tan\frac{\pi}{3}\right) - \log(1 + 0)\right]\)

\(= \frac{1}{2}\log(2 + \sqrt{3})\).