Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 30

Answer

$-2 \csc (\frac{v-\pi}{2})+C$

Work Step by Step

Let u=$\csc (\frac{v-\pi}{2})=>\frac{-1}{2} \csc(\frac{v-\pi}{2})\cot (\frac{v-\pi}{2})dv$ $=>-2du=\csc(\frac{v-\pi}{2}) \cot(\frac{v-\pi}{2})dv$ =$\csc(\frac{v-\pi}{2}) \cot(\frac{v-\pi}{2})$ =$\int -2du$ =$-2u+C$ =$-2 \csc (\frac{v-\pi}{2})+C$
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