Chapter 4 - Applications of the Derivative - Review Exercises - Page 332: 78

$${\sin ^{ - 1}}x + C$$

$$\eqalign{ & \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \cr & {\text{by the differentiation formula }}\frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x} \right] = \frac{1}{{\sqrt {1 - {x^2}} }} \cr & d\left[ {{{\sin }^{ - 1}}x} \right] = \frac{1}{{\sqrt {1 - {x^2}} }}dx \cr & {\text{then}}{\text{, we conclude that}} \cr & \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = {\sin ^{ - 1}}x + C \cr} $$