University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 36



Work Step by Step

$$A=\int^{\pi/12}_06\tan3xdx=\int^{\pi/12}_0\frac{6\sin3x}{\cos3x}dx$$ We set $u=\cos3x$, which means $$du=-3\sin3xdx=-\frac{1}{2}\times(6\sin3xdx)$$ $$6\sin3xdx=-2du$$ For $x=\pi/12$, we have $$u=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$$ For $x=0$, we have $$u=\cos0=1$$ Therefore, $$A=-2\int^{\sqrt2/2}_1\frac{1}{u}du=-2\ln|u|\Big]^{\sqrt2/2}_1$$ $$A=-2\Big(\ln\frac{\sqrt2}{2}-\ln1\Big)$$ $$A=-2\ln\frac{\sqrt2}{2}$$ $$A=\ln\Big(\frac{\sqrt2}{2}\Big)^{-2}=\ln\frac{4}{2}$$ $$A=\ln2$$
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