Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.9 Antiderivatives - 3.9 Exercises - Page 282: 17

Answer

$$H(\theta) =-2\cos \theta -\tan\theta+C $$

Work Step by Step

Given $$h(\theta) = 2\sin \theta -\sec^2\theta$$ Then by using table 2 if $f(x)=\sin x\ \to\ \ F(x) =-\cos +C$ and if $f(x)=\sec^2x\ \to\ \ F(x) =\tan x +C$ Hence \begin{align*} H(\theta) &=-2\cos \theta -\tan\theta+C \end{align*} To check \begin{align*} H'(\theta) &= 2\sin \theta -\sec^2\theta\\ &=h(\theta) \end{align*}
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