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: 37

Answer

$$\int^{\pi/4}_0\frac{1+\cos^2\theta}{\cos^2\theta}d\theta=1+\frac{\pi}{4}$$

Work Step by Step

$$A=\int^{\pi/4}_0\frac{1+\cos^2\theta}{\cos^2\theta}d\theta$$ $$A=\int^{\pi/4}_0(\frac{1}{\cos^2\theta}+1)d\theta$$ $$A=\int^{\pi/4}_0(\sec^2\theta+1)d\theta$$ According to Table 1, we have $$\int \sec^2 xdx=\tan x+C$$ $$\int kdx=kx+C$$ Therefore, $$A=(\tan\theta+\theta)\Bigg]^{\pi/4}_0$$ $$A=(\tan\frac{\pi}{4}+\frac{\pi}{4})-(\tan0+0)$$ $$A=(1+\frac{\pi}{4})-(0+0)$$ $$A=1+\frac{\pi}{4}$$
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