Calculus (3rd Edition)

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

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 28

Answer

$$\int \sec^2(12-25\theta) d\theta= -\frac{1}{25} \tan (12-25\theta)+c.$$

Work Step by Step

Recall that $(\tan x)'=\sec^2 x$. Let $ u=12-25\theta $, then $ du=-25d\theta $. $$\int \sec^2(12-25\theta) d\theta=-\frac{1}{25}\int \sec^2udu\\ =-\frac{1}{25} \tan u+c=-\frac{1}{25} \tan (12-25\theta)+c.$$
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