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: 31


$$\int^{\pi/2}_{0}7^{\cos t}\sin tdt=\frac{6}{\ln7}$$

Work Step by Step

$$A=\int^{\pi/2}_{0}7^{\cos t}\sin tdt$$ We set $\cos t=u$, which means $$-\sin tdt=du$$ $$\sin tdt=-du$$ - For $t=\pi/2$, we have $u=\cos(\pi/2)=0$ - For $t=0$, we have $u=\cos0=1$ Therefore, $$A=-\int^{0}_{1}7^udu$$ $$A=-\frac{7^u}{\ln7}\Big]^{0}_{1}$$ $$A=-\frac{1}{\ln7}(7^0-7^1)$$ $$A=-\frac{-6}{\ln7}=\frac{6}{\ln7}$$
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