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: 44

Answer

$$-2\ln|\cos \sqrt x|+c.$$

Work Step by Step

Let $u=\sqrt x$; then $du=\frac{1}{2\sqrt x}dx$. Hence, we have $$\int \frac{ \tan\sqrt x}{\sqrt x} d x=2\int \tan u \ du \\ =2\int \frac{\sin u}{\cos u}du=-2\int \frac{d\cos u}{\cos u}\\ =-2\ln|\cos u|+c=-2\ln|\cos \sqrt x|+c.$$
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