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: 32


$$\int^{\pi/4}_{0}\Big(\frac{1}{3}\Big)^{\tan t}\sec^2 tdt=\frac{2}{3\ln3}$$

Work Step by Step

$$A=\int^{\pi/4}_{0}\Big(\frac{1}{3}\Big)^{\tan t}\sec^2 tdt$$ We set $\tan t=u$, which means $$\sec^2 tdt=du$$ - For $t=\pi/4$, we have $u=\tan(\pi/4)=1$ - For $t=0$, we have $u=\tan0=0$ Therefore, $$A=\int^{1}_{0}\Big(\frac{1}{3}\Big)^udu$$ $$A=\frac{\Big(\frac{1}{3}\Big)^u}{\ln\frac{1}{3}}\Big]^{1}_{0}$$ $$A=\frac{1}{\ln\frac{1}{3}}(\frac{1}{3}-1)$$ $$A=-\frac{2}{3\ln\frac{1}{3}}$$ We can further rewrite $\ln(1/3)=\ln1-\ln3=-\ln3$ Thus, $$A=\frac{2}{3\ln3}$$
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