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: 20


$$\frac{t^{2}-25}{25 t}+C$$

Work Step by Step

Given $$\int \frac{d t}{t^{2} \sqrt{t^{2}-25}}$$ Let $$ t=5 \sec \theta,\ \ \ \ dt = 5\sec \theta \tan \theta d \theta $$ \begin{aligned} \int \frac{d t}{t^{2} \sqrt{t^{2}-25}} &=\int \frac{5 \sec \theta \tan \theta d \theta}{(5 \sec \theta)^{2} \sqrt{(5 \sec \theta)^{2}-25}} \\ &=\int \frac{5 \sec \theta \tan \theta d \theta}{25 \sec ^{2} \theta \sqrt{25 \sec ^{2} \theta-25}} \\ &=\int \frac{\tan \theta d \theta}{5 \sec \theta \cdot 5 \sqrt{\sec ^{2} \theta-1}} \\ &=\frac{1}{25} \int \frac{\tan \theta d \theta}{\sec \theta} \\ &=\frac{1}{25} \int \frac{1}{\sec \theta} d \theta \\ &=\frac{1}{25} \int \cos \theta d \theta \\ &=\frac{1}{25} \sin \theta+C\\ &=\frac{t^{2}-25}{25 t}+C \end{aligned}
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