University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 45


$$\int\cos\sqrt xdx=2(\sqrt x\sin \sqrt x+\cos\sqrt x)+C$$

Work Step by Step

$$A=\int\cos\sqrt xdx$$ We set $a=\sqrt x$, which means $$da=\frac{1}{2\sqrt x}dx=\frac{1}{2a}dx$$ $$dx=2ada$$ Therefore, $$A=2\int a\cos ada$$ Next, we do integration by parts. Take $u=a$ and $dv=\cos ada$ We would have $du=da$ and $v=\sin a$ Applying the formula $\int udv=uv-\int vdu$, we have $$A=2(a\sin a-\int \sin ada)$$ $$A=2(a\sin a+\cos a)+C$$ $$A=2(\sqrt x\sin \sqrt x+\cos\sqrt x)+C$$
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