Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 44



Work Step by Step

We have $$ \int_{0}^{2\pi}f(x)dx=\int_{0}^{\pi}f(x)dx+\int_{\pi}^{2\pi}f(x)dx\\ =\int_{0}^{\pi}\cos x dx+\int_{\pi}^{2\pi}\cos x -\sin 2x dx\\ =\sin x|_{0}^{\pi}+(\sin x+\frac{1}{2}\cos 2x)|_{\pi}^{2\pi} \\ =0+\frac{1}{2}(1-1)=0 $$
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