## Calculus: Early Transcendentals (2nd Edition)

$$\frac{{\sqrt 3 - 1}}{4}$$
\eqalign{ & \int_1^{\sqrt 2 } {\frac{{dx}}{{{x^2}\sqrt {4 - {x^2}} }}} \cr & x = 2\sin \theta ,{\text{ }}dx = 2\cos \theta d\theta \cr & {\text{new limits of integration}} \cr & \theta = {\sin ^{ - 1}}\left( {x/2} \right) \cr & x = \sqrt 2 ,{\text{ }}\theta = \pi /4 \cr & x = 1,{\text{ }}\theta = \pi /6 \cr & {\text{we have}} \cr & \int_1^{\sqrt 2 } {\frac{{dx}}{{{x^2}\sqrt {4 - {x^2}} }}} = \int_{\pi /6}^{\pi /4} {\frac{{2\cos \theta d\theta }}{{4{{\sin }^2}\theta \sqrt {4 - 4{{\sin }^2}\theta } }}} \cr & = \int_{\pi /6}^{\pi /4} {\frac{{2\cos \theta d\theta }}{{4{{\sin }^2}\theta \left( 2 \right)\sqrt {1 - {{\sin }^2}\theta } }}} \cr & {\text{pythagorean identity}} \cr & = \frac{1}{4}\int_{\pi /6}^{\pi /4} {\frac{{\cos \theta d\theta }}{{{{\sin }^2}\theta \sqrt {{{\cos }^2}\theta } }}} = \frac{1}{4}\int_{\pi /6}^{\pi /4} {{{\csc }^2}\theta } d\theta \cr & {\text{integrating}} \cr & = - \frac{1}{4}\left[ {\cot \theta } \right]_{\pi /6}^{\pi /4} \cr & {\text{evaluate limits}} \cr & = - \frac{1}{4}\left[ {\cot \left( {\frac{\pi }{4}} \right) - \cot \left( {\frac{\pi }{6}} \right)} \right] \cr & {\text{simplify}} \cr & = - \frac{1}{4}\left[ {1 - \sqrt 3 } \right] \cr & = \frac{{\sqrt 3 - 1}}{4} \cr}