University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 62


$$\int 2^{\tan x}\sec^2xdx=\frac{2^{\tan x}}{\ln2}+C$$

Work Step by Step

$$A=\int 2^{\tan x}\sec^2xdx$$ We set $a=\tan x$, which means $$da=\sec^2xdx$$ Therefore, $$A=\int 2^ada$$ $$A=\frac{2^a}{\ln2}+C$$ $$A=\frac{2^{\tan x}}{\ln2}+C$$
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