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: 16


$$\frac{1}{16} \frac{t}{\sqrt{16-t^{2}}}+C$$

Work Step by Step

\begin{aligned} \int \frac{d t}{\left(16-t^{2}\right)^{\frac{3}{2}}} &=\int \frac{4 \cos \theta d \theta}{\left(16-(4 \sin \theta)^{2}\right)^{\frac{3}{2}}} \\ &=\int \frac{4 \cos \theta d \theta}{(\sqrt{16-16 \sin ^{2} \theta})^{3}} \\ &=\int \frac{4 \cos \theta d \theta}{(4 \sqrt{1-\sin ^{2} \theta})^{3}} \\ &=\frac{1}{16} \int \frac{\cos \theta d \theta}{(\sqrt{\cos ^{2} \theta})^{3}} \\ &=\frac{1}{16} \int \frac{\cos \theta d \theta}{(\cos \theta)^{3}}\\ &=\frac{1}{16}\int \sec^2 \theta d\theta\\ &=\frac{1}{16}\tan \theta +C\\ &=\frac{1}{16} \frac{t}{\sqrt{16-t^{2}}}+C \end{aligned}
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