## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{{\sqrt 2 }}{\sin ^{ - 1}}\sqrt 2 x + C$$
\eqalign{ & \int {\frac{{dx}}{{\sqrt {1 - 2{x^2}} }}} \cr & {\text{substitute }}x = \frac{1}{{\sqrt 2 }}sin\theta ,{\text{ }}dx = \frac{1}{{\sqrt 2 }}\cos \theta d\theta \cr & \int {\frac{{dx}}{{\sqrt {1 - 2{x^2}} }}} = \int {\frac{{\left( {1/\sqrt 2 } \right)\cos \theta d\theta }}{{\sqrt {1 - 2{{\left( {\frac{1}{{\sqrt 2 }}\sin \theta } \right)}^2}} }}} = \frac{1}{{\sqrt 2 }}\int {\frac{{\cos \theta d\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}} \cr & {\text{pythagorean identity}} \cr & = \frac{1}{{\sqrt 2 }}\int {\frac{{\cos \theta d\theta }}{{\sqrt {{{\cos }^2}\theta } }}} = \frac{1}{{\sqrt 2 }}\int {\frac{{\cos \theta d\theta }}{{\cos \theta }}} = \frac{1}{{\sqrt 2 }}\int {d\theta } \cr & {\text{integrating}} \cr & = \frac{1}{{\sqrt 2 }}\theta + C \cr & x = \frac{1}{{\sqrt 2 }}sin\theta \therefore {\text{ }}\theta = {\sin ^{ - 1}}\sqrt 2 x \cr & = \frac{1}{{\sqrt 2 }}{\sin ^{ - 1}}\sqrt 2 x + C \cr}