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: 41


$-\dfrac{\csc \theta}{2}+C$

Work Step by Step

As we are given that $\int (\dfrac{1}{2} \csc \theta \cot \theta) d\theta$ Thus,we have $\int (\dfrac{1}{2} \csc \theta \cot \theta) d\theta= \dfrac{1}{2}\int ( \csc \theta \cot \theta) d\theta$ or, $=-\dfrac{\csc \theta}{2}+C$
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