Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 239: 45


$-\dfrac{1}{2} \cos 2x +\cot x+C$

Work Step by Step

Solve. $\int (\sin 2x ) dx - \int (\csc^2 x) dx$ Thus, we have $\int (\sin 2x ) dx - \int (\csc^2 x) dx=-\dfrac{1}{2} \cos 2x -(-\cot x)+C$ or, $=-\dfrac{1}{2} \cos 2x +\cot x+C$
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