Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 110


$$-\frac{1}{2\ln 5} 5^{-2\sin x}+c $$

Work Step by Step

Let $u=\sin x$, then $du=\cos x \ dx$. We have $$\int \cos x \ 5^{-2\sin x}dx=\int 5^{-2u}du=-\frac{1}{2\ln 5}\int (-2\ln 5) \ 5^{-2u}du\\ =-\frac{1}{2\ln 5} 5^{-2u}+c=-\frac{1}{2\ln 5} 5^{-2\sin x}+c $$
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