Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 49


$-2 sech \sqrt t +C$

Work Step by Step

As we are given that $\int \dfrac{sech \sqrt t \tanh \sqrt t dt}{\sqrt t}$ Re-write:$\int \dfrac{sech \sqrt t \tanh \sqrt t dt}{\sqrt t}=2 \int (sech \sqrt t \tanh \sqrt t)\dfrac{ dt}{ 2\sqrt t}$ Now, consider $\sqrt t =u$ and $du= \dfrac{ dt}{ 2\sqrt t}$ This implies, $2 \int (sech \sqrt t \tanh \sqrt t)\dfrac{ dt}{ 2\sqrt t}=2 \int (sech u \tanh u) du$ or, $=-2 (sech u) +C$ or, $=-2 sech \sqrt t +C$
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.