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: 29

Answer

$\frac{1}{8}\tan^8 x+\frac{1}{3}\tan^6 x+\frac{1}{4}\tan^4 x+C$

Work Step by Step

$\int\tan^3 x\sec^6 x\ dx$ Since the power of $\sec x$ is even, save a factor of $\sec^2 x$ and express the remaining factors in terms of $\tan x$: $=\int \tan^3 x\sec^4 x\sec^2 x\ dx$ $=\int\tan^3 x(\sec^2 x)^2\sec^2 x\ dx$ $=\int\tan^3 x(1+\tan^2 x)^2\sec^2 x\ dx$ Let $u=\tan x$. The $du=\sec^2 x\ dx$. $=\int u^3(1+u^2)^2\ du$ $=\int u^3(1+2u^2+u^4)\ du$ $=\int (u^7+2u^5+u^3)\ du$ $=\frac{1}{8}u^8+\frac{2}{6}u^6+\frac{1}{4}u^4+C$ $=\boxed{\frac{1}{8}\tan^8 x+\frac{1}{3}\tan^6 x+\frac{1}{4}\tan^4 x+C}$
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