Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises - Page 525: 34


$$\int \frac{sin(\phi)}{cos^{3}(\phi)}d\phi=\frac{1}{2}tan^{2}\phi+C$$

Work Step by Step

$$\int \frac{sin(\phi)}{cos^{3}(\phi)}d\phi=\int tan\phi\frac{1}{cos^{2}\phi}d\phi$$ $$=\int tan\phi\,sec^{2}\phi\,d\phi$$ $$=\int tan\phi\,d(tan\phi)$$ $$=\frac{1}{2}tan^{2}\phi+C$$ $\frac{1}{2}tan^{2}\phi +C= \frac{1}{2}sec^{2}\phi-\frac{1}{2}+C=\frac{1}{2}sec^{2}\phi+C_{1}$ The two answers are the same.
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