Answer
$-sin(\frac{1}{t}-1)+C$
Work Step by Step
Let u=$\frac{1}{t}-1=t^{-1}-1=>du=-t^{-2}=>-du-\frac{1}{t^5}dt$
$\frac{1}{t^2}cos(\frac{1}{t}-1)dt$
=$\int (cos u)(-du)$
=$-\int \cos udu=2\sin u+C$
=$-sin(\frac{1}{t}-1)+C$
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 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 33
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