Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 6



Work Step by Step

Since $x=\sin \theta $, then $dx=\cos \theta d\theta$, and when $x :0\to 1/2$ then $\theta :0\to \pi/6$; hence, we have $$ \int_{0}^{1 / 2} \frac{x^{2}}{\sqrt{1-x^{2}}} d x=\int_{0}^{\pi / 6} \frac{\sin^2 \theta }{\sqrt{1-\sin^{2}\theta}} d \theta\\ =\int_{0}^{\pi / 6} \frac{\sin^2 \theta }{\sqrt{1-\sin^{2}\theta}} \cos \theta d \theta\\ =\int_{0}^{\pi / 6}\sin^2\theta d\theta=\frac{1}{2}\int_{0}^{\pi / 6}(1-\cos 2\theta ) d\theta\\ =\frac{1}{2}(\theta -\frac{1}{2} \sin2\theta)|_{0}^{\pi / 6}= =\frac{1}{2}\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right). $$
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