Answer
$$\int 2^{\tan x}\sec^2xdx=\frac{2^{\tan x}}{\ln2}+C$$
Work Step by Step
$$A=\int 2^{\tan x}\sec^2xdx$$
We set $a=\tan x$, which means $$da=\sec^2xdx$$
Therefore, $$A=\int 2^ada$$ $$A=\frac{2^a}{\ln2}+C$$ $$A=\frac{2^{\tan x}}{\ln2}+C$$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 5 - Practice Exercises - Page 343: 62
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