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: 5


$$\frac{8}{\sqrt{5}}(\sin^{-1}(\frac{\sqrt{5} x}{4})+\frac{\sqrt{5} x}{4}\frac{\sqrt{16-5x^2}}{4}) +c $$

Work Step by Step

Since $ x=\frac{4}{\sqrt{5}}\sin \theta $, then $ dx=\frac{4}{\sqrt{5}}\cos \theta d \theta $ and hence $$\int \sqrt{16-5x^2} dx=\frac{4}{\sqrt{5}}\int \sqrt{16-16 \sin^2 \theta} \cos \theta d \theta\\ =\frac{16}{\sqrt{5}}\int \cos^2 \theta d \theta\\ =\frac{16}{\sqrt{5}}\int \cos^2 \theta d \theta\\ =\frac{8}{\sqrt{5}}\int 1+ \cos2 \theta d \theta\\ =\frac{8}{\sqrt{5}}(\theta+\frac{1}{2} \sin2 \theta )+c\\ =\frac{8}{\sqrt{5}}(\theta+ \sin \theta \cos \theta)+c\\ =\frac{8}{\sqrt{5}}(\sin^{-1}(\frac{\sqrt{5} x}{4})+\frac{\sqrt{5} x}{4}\frac{\sqrt{16-5x^2}}{4}) +c $$ Note that we used the identity $\sin 2\theta=2\cos \theta \sin \theta$.
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