Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.2 The Indefinite Integral - Exercises Set 4.2 - Page 279: 31


$$\tan x-\sec x+c$$

Work Step by Step

Given$$\int \frac{dx}{1+\sin x}$$ Since \begin{align*} \int \frac{dx}{1+\sin x}\frac{1-\sin x}{1-\sin x} &=\int \frac{(1-\sin x)dx}{1-\sin^2 x}\\ &=\int \frac{(1-\sin x)dx}{\cos^2 x}\\ &=\int [\sec^2 x-\sec x\tan x]dx\\ &=\tan x-\sec x+c \end{align*}
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