University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 45



Work Step by Step

$$A=\int(2\theta+1+2\cos(2\theta+1))d\theta$$ We set $a=2\theta+1$, then $$da=2d\theta$$ $$d\theta=\frac{1}{2}da$$ Therefore, $$A=\frac{1}{2}\int (a+2\cos a)da$$ $$A=\frac{1}{2}\Big(\int ada+2\int\cos ada\Big)$$ $$A=\frac{1}{2}\Big(\frac{a^2}{2}+2\sin a\Big)+C$$ $$A=\frac{a^2}{4}+\sin a+C$$ $$A=\frac{(2\theta+1)^2}{4}+\sin(2\theta+1)+C$$
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