University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 14


$-\dfrac{[\ln (\cos x)]^2}{2}+c$

Work Step by Step

Solve $\int \tan x \ln (\cos x) dx$ Let us $p=\ln (\cos x)$ and $dp=- \tan x dx$ This implies $\int \tan x \ln (\cos x) dx=-\int p dp $ $= -\dfrac{p^2}{2}+c$ $= -\dfrac{[\ln (\cos x)]^2}{2}+c$ Hence, $\int \tan x \ln (\cos x) dx=-\dfrac{[\ln (\cos x)]^2}{2}+c$
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