Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 42


$$\int \sin \sqrt x d x=2( \sin \sqrt x - \sqrt x \cos \sqrt x )+c$$

Work Step by Step

Let $u=\sqrt x$; then $du=\frac{1}{2\sqrt x}dx$. Hence, we have $$\int \sin \sqrt x d x =2 \int u \sin u \ du.$$ Now we use integration by parts as follows: $$ \int u \sin u \ du=-u \cos u +\int \cos u \ du=-u \cos u + \sin u +c.$$ Hence $$\int \sin \sqrt x d x=2( \sin \sqrt x - \sqrt x \cos \sqrt x )+c$$
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