Explanation Given information: The integral is ∫ e x − e − x e x + e − x d x Calculation: Take ∫ e x − e − x e x + e − x d x (1) Substituting e x + e − x = w ⇒ ( e x − e − x ) d x = d w Using these in

Applied Calculus, 6e WileyPLUS + Loose-leaf

6th Edition

ISBN: 9781119408901

Author: Deborah Hughes-Hallett, Andrew M. Gleason, Patti Frazer Lock, Daniel E. Flath

Publisher: Wiley (WileyPLUS Products)

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Chapter 6, Problem 57SYU

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True or false:

The substitution w=ex+e−x converts the integral ∫ex−e−xex+e−xdx into ∫1wdw.

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Chapter 6, Problem 56SYU

Chapter 6, Problem 58SYU

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Chapter 6 Solutions

Applied Calculus, 6e WileyPLUS + Loose-leaf

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True or false: The substitution w = e x + e − x converts the integral ∫ e x − e − x e x + e − x d x into ∫ 1 w d w .

Solution Summary: The author explains that the substitution w=ex+

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True or false:

The substitution w=ex+e−x converts the integral ∫ex−e−xex+e−xdx into ∫1wdw.

Want to see the full answer?

Chapter 6 Solutions

True or false: The substitution w = e x + e − x converts the integral ∫ e x − e − x e x + e − x d x into ∫ 1 w d w .

Solution Summary: The author explains that the substitution w=ex+

Concept explainers

Videos

True or false: The substitution w=ex+e−x converts the integral ∫ex−e−xex+e−xdx into ∫1wdw.

Want to see the full answer?

Chapter 6 Solutions

True or false:

The substitution w=ex+e−x converts the integral ∫ex−e−xex+e−xdx into ∫1wdw.