Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises: 20

Answer

$ x tan (x) + ln|cos (x)| - x^2/2 + C $

Work Step by Step

Using integration by parts: u = x so u' = 1 v' = $tan^2 (x) = sec^2 (x) -1 $ (Trigonometry identity) v = $ \int sec^2 (x) -1 $ dx = $ tan(x) - x $ Now let's plug those parts into the equation: $\int x tan^2 (x) dx = uv - \int (u'v)$ =$ x(tan(x) - x) - \int (tan(x) - x)dx $ = $ x tan (x) - x^2 + ln|cos (x)| + (x^2)/2 + C $ = $ x tan (x) + ln|cos (x)| - x^2/2 + C $
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