University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Practice Exercises - Page 279: 112


$\dfrac{1}{2} x+\dfrac{1}{2} \sin x+C$

Work Step by Step

Calculate the anti-derivative. Since, we know $\cos^2 x=\dfrac{1+\cos x}{2}$ and $\int \cos x dx=\sin x$ Then $\int \cos^2 (x/2) dx= \int \dfrac{1+\cos x}{2} dx$ or, $=\dfrac{1}{2} x+ \dfrac{1}{2} \int \cos x dx$ Thus, $\int \cos^2 (x/2) dx=\dfrac{1}{2} x+\dfrac{1}{2} \sin x+C$
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