Answer
$\dfrac{2}{3}\tan^3t+C$
Work Step by Step
Let $I=\int 2\sin^2 t\sec^4t dt=2\int\tan^2t\sec^2t \space dt$
Suppose $u=\tan t \implies du=\sec^2tdt$
Therefore, $I=2\int u^2du \\=\dfrac{2u^3}{3}+C \\=\dfrac{2}{3}\tan^3t+C$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 8: Techniques of Integration - Section 8.6 - Integral Tables and Computer Algebra Systems - Exercises 8.6 - Page 481: 44
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