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: 21

Answer

$\frac{1}{3}\sec^3 x+C$

Work Step by Step

$\int\tan x\sec^3 x\ dx$ Since the power of tangent is odd, save a factor of $\sec x\tan x$. $=\int\sec^2 x\sec x\tan x\ dx$ Let $u=\sec x$. Then $du=\sec x\tan x\ dx$. $=\int u^2 du$ $=\frac{1}{3}u^3+C$ $=\boxed{\frac{1}{3}\sec^3 x+C}$
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