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: 15

Answer

$$\int(2+\tan^2\theta)d\theta=\theta+\tan\theta+C$$

Work Step by Step

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