Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 252: 37


$\frac{3}{4} \tan(4 x) -\frac{5}{4} \sec (4 x) +c.$

Work Step by Step

Since $(\tan 4 x)'=4\sec^2(4x)$ and $(\sec^2(4x))'=4\sec^2(4x)\tan(4x)$, we have $$ \int \sec 4 x(3 \sec 4 x-5 \tan 4 x) d x=\int3 \sec^2(4 x) -5\sec (4 x)\tan (4 x) d x\\ =\frac{3}{4} \tan(4 x) -\frac{5}{4} \sec (4 x) +c. $$
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