Answer
$${\text{True}}$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \cr
& {\text{Let }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr
& {\text{Substituting}} \cr
& \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = \int {\frac{{\cos \theta d\theta }}{{\sqrt {1 - {{\left( {\sin \theta } \right)}^2}} }}} \cr
& = \int {\frac{{\cos \theta d\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}} \cr
& {\text{Simplifying}} \cr
& = \int {\frac{{\cos \theta d\theta }}{{\cos \theta }}} \cr
& = \int {d\theta } \cr
& {\text{The statement is True}} \cr} $$
