Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises: 16

Answer

$$\int\sec t(\sec t+\tan t)dt=\tan t+\sec t+C$$

Work Step by Step

$$A=\int\sec t(\sec t+\tan t)dt$$ $$A=\int(\sec^2 t+\sec t\tan t)dt$$ From Table 1, $$\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$$ Therefore, $$A=\int(\sec^2 t)dt+\int(\sec t\tan t)dt$$ From Table 1, we also get the followings $$\int(\sec^2 x)dx=\tan x+C$$ $$\int(\sec x\tan x)dx=\sec x+C$$ Therefore, $$A=\tan t+\sec t+C$$
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