[-InnoXenT-]

let $x = \sin{\theta}$ ---> $dx = \cos{\theta} \, d\theta$

$$\int_{0}^{\sin{\alpha}} \frac{dx}{\sqrt{1-x^2}} = \int_{0}^{\alpha} \frac{\cos{\theta} \, d\theta}{\sqrt{1-\sin^2{\theta}}}$$

$$= \left[\theta \right]_{0}^{\alpha} $$

$$= \alpha$$