Explanation

(b)
Let $I = \int_{\alpha}^{\beta}\sqrt{\frac{x - \alpha}{\beta - x}dx}$
Put $x = \alpha\cos^{2}{\theta + \beta\sin^{2}\theta}$,
$\therefore x - \alpha = (\beta - \alpha)\sin^{2}{\theta,\ \beta - x = (\beta - \alpha)\cos^{2}\theta}$
and $dx = 2(\beta - \alpha)\sin{\theta\cos{\theta d\theta}}$
$\therefore I = \int_{0}^{\pi/2}\sqrt{\frac{(\beta - \alpha)\sin^{2}\theta}{(\beta - \alpha)\cos^{2}\theta} \times 2(\beta - \alpha)\sin{\theta\cos\theta}}d\theta$
$= (\beta - \alpha)\int_{0}^{\pi/2}{(1 - cos2\theta)d\theta}$
$= (\beta - \alpha)\left\lbrack \theta - \frac{\sin{2\theta}}{2} \right\rbrack_{0}^{\pi/2} = (\beta - \alpha).\frac{\pi}{2}$