Calculus 8th Edition

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

Chapter 4 - Integrals - 4.4 Indefinite Integrals and the Net Change Theorem - 4.4 Exercises - Page 336: 3

Answer

Verify $\int\tan^2{x}dx=\tan{x}-x+C$ We need to take the derivative of the right side of our equation and verify that it equals the expression inside of our integral. \begin{equation*} \frac{d}{dx}\left(\tan{x}-x+C\right)=\frac{d}{dx}\left(\tan{x}\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(C\right) \end{equation*} \begin{equation*} =\left(\sec^2{x}\right)-\left(1\right)+\left(0\right) \end{equation*} \begin{equation*} =\tan^2{x} \end{equation*} as desired.

Work Step by Step

Verify $\int\tan^2{x}dx=\tan{x}-x+C$ We need to take the derivative of the right side of our equation and verify that it equals the expression inside of our integral. \begin{equation*} \frac{d}{dx}\left(\tan{x}-x+C\right)=\frac{d}{dx}\left(\tan{x}\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(C\right) \end{equation*} \begin{equation*} =\left(\sec^2{x}\right)-\left(1\right)+\left(0\right) \end{equation*} \begin{equation*} =\tan^2{x} \end{equation*} as desired.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.