Chapter 7 - Section 7.3 - Trigonometric Substitution - 7.3 Exercises - Page 505: 22

$\frac{\pi }{{24}}$

$$\eqalign{ & \int_{1/4}^{\sqrt 3 /4} {\sqrt {1 - 4{x^2}} } dx \cr & {\text{Let}} \cr & \sin \theta = 2x,{\text{ }}x = \frac{1}{2}\sin \theta ,{\text{ }}dx = \frac{1}{2}\cos \theta d\theta \cr & \sin \theta = 2x \to \theta = \arcsin 2x \cr & \sqrt {1 - 4{x^2}} = \cos \theta \cr & {\text{The new limits of integration are:}} \cr & x = \frac{{\sqrt 3 }}{4} \to \theta = \arcsin 2\left( {\frac{{\sqrt 3 }}{4}} \right) = \frac{\pi }{3} \cr & x = \frac{1}{4} \to \theta = \arcsin 2\left( {\frac{1}{4}} \right) = \frac{\pi }{6} \cr & {\text{Substituting}}{\text{, we obtain}} \cr & \int_{1/4}^{\sqrt 3 /4} {\sqrt {1 - 4{x^2}} } dx = \int_{\pi /6}^{\pi /3} {\cos \theta \left( {\frac{1}{2}\cos \theta } \right)d\theta } \cr & = \frac{1}{2}\int_{\pi /6}^{\pi /3} {{{\cos }^2}\theta d\theta } \cr & = \frac{1}{2}\int_{\pi /6}^{\pi /3} {\left( {\frac{{1 + \cos 2\theta }}{2}} \right)} d\theta \cr & = \frac{1}{4}\int_{\pi /6}^{\pi /3} {\left( {1 + \cos 2\theta } \right)} d\theta \cr & {\text{Integrating}} \cr & = \frac{1}{4}\left[ {\theta + \frac{1}{2}\sin 2\theta } \right]_{\pi /6}^{\pi /3} \cr & = \frac{1}{4}\left[ {\theta + \frac{1}{2}\left( {2\sin \theta \cos \theta } \right)} \right]_{\pi /6}^{\pi /3} \cr & = \frac{1}{4}\left[ {\theta + \sin \theta \cos \theta } \right]_{\pi /6}^{\pi /3} \cr & {\text{Evaluating}} \cr & = \frac{1}{4}\left[ {\frac{\pi }{3} + \sin \left( {\frac{\pi }{3}} \right)\cos \left( {\frac{\pi }{3}} \right)} \right] - \frac{1}{4}\left[ {\frac{\pi }{6} + \sin \left( {\frac{\pi }{6}} \right)\cos \left( {\frac{\pi }{6}} \right)} \right] \cr & = \frac{1}{4}\left[ {\frac{\pi }{3} + \frac{{\sqrt 3 }}{4}} \right] - \frac{1}{4}\left[ {\frac{\pi }{6} + \frac{{\sqrt 3 }}{4}} \right] \cr & = \frac{\pi }{{12}} - \frac{\pi }{{24}} \cr & = \frac{\pi }{{24}} \cr} $$