Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises - Page 499: 55

Answer

$\csc x + \cot x + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{dx}}{{\cos x - 1}}} \cr & {\text{Multiplty the numerator and denominator by the conjugate of }} \cr & {\text{the denominator}} \cr & = \int {\frac{1}{{\cos x - 1}}} \times \frac{{\cos x + 1}}{{\cos x + 1}}dx \cr & = \int {\frac{{\cos x + 1}}{{{{\cos }^2}x - 1}}} dx \cr & {\text{Use the pythagorean identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr & = \int {\frac{{\cos x + 1}}{{ - {{\sin }^2}x}}} dx \cr & = - \int {\left( {\frac{{\cos x}}{{{{\sin }^2}x}} + \frac{1}{{{{\sin }^2}x}}} \right)} dx \cr & = - \int {\left( {\csc x\cot x + {{\csc }^2}x} \right)} dx \cr & {\text{Integrating}} \cr & = - \left( { - \csc x - \cot x} \right) + C \cr & = \csc x + \cot x + C \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.