Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 7 - Principles Of Integral Evaluation - 7.2 Integration By Parts - Exercises Set 7.2 - Page 498: 23

Answer

$$x\tan x + \ln \left| {\cos x} \right| + C$$

Work Step by Step

$$\eqalign{ & \int {x{{\sec }^2}x} dx \cr & {\text{substitute }}u = x,{\text{ }}du = dx \cr & dv = {\sec ^2}xdx,{\text{ }}v = \tan x \cr & {\text{ integration by parts}} \cr & \int {udv} = uv - \int {vdu} \cr & {\text{, we have}} \cr & \int {x{{\sec }^2}x} dx = x\tan x - \int {\tan xdx} \cr & {\text{trigonometric identity }}\tan x = \frac{{\sin x}}{{\cos x}} \cr & \int {x{{\sec }^2}x} dx = x\tan x - \int {\frac{{\sin x}}{{\cos x}}dx} \cr & \int {x{{\sec }^2}x} dx = x\tan x + \int {\frac{{ - \sin x}}{{\cos x}}dx} \cr & {\text{find antiderivative}} \cr & \int {x{{\sec }^2}x} dx = x\tan x + \ln \left| {\cos x} \right| + C \cr} $$
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