Answer
$${\sin ^{ - 1}}x + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \cr
& {\text{Using Trigonometric Substitutions}} \cr
& {\text{ }}x = \sin \theta,{\text{ }}dx = \cos \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = \int {\frac{{\cos \theta d\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}} \cr
& {\text{use the fundamental identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr
& = \int {\frac{{\cos \theta d\theta }}{{\sqrt {{{\cos }^2}\theta } }}} \cr
& {\text{simplifying}} \cr
& = \int {\frac{{\cos \theta d\theta }}{{\cos \theta }}} \cr
& = \int {d\theta } \cr
& {\text{integrate}} \cr
& = \theta + C \cr
& {\text{write in terms of }}x,{\text{ }}x = \sin \theta \,\,\,\, \Rightarrow \,\,\,\theta = {\sin ^{ - 1}}x \cr
& = {\sin ^{ - 1}}x + C \cr} $$
