Integration and inverse trig functions

Click For Summary

Homework Help Overview

The discussion revolves around the integration of hyperbolic functions, specifically the integral of sinh(x) divided by (1 + cosh(x)). Participants express confusion regarding the appropriate approach to tackle this integral and whether to utilize any identities.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants suggest various approaches, including the use of identities and variable changes. There is a mention of a potential simplification involving tanh(x/2), and some participants question the correctness of certain equalities presented in the discussion.

Discussion Status

The discussion is active, with participants offering different perspectives on how to approach the integral. Some guidance has been provided regarding variable substitution and potential simplifications, though there is no explicit consensus on the best method to proceed.

Contextual Notes

There is a recognition that the functions in question are hyperbolic rather than inverse trigonometric functions, which may influence the approach taken by participants. Additionally, some participants express uncertainty about the correctness of their statements, indicating a reflective and questioning atmosphere.

mavsqueen06
Messages
7
Reaction score
0

Homework Statement


getting confused with integration of trig functions.
I am finding the integral of sinhx/1+coshx and I'm not sure how to start. should i use an identity?

help is appreciated!


Homework Equations


possibly an identity of some sort?

The Attempt at a Solution


?
 
Physics news on Phys.org
Well the obvious chamge would be;

[tex]\frac{sinh(x)}{1+cosh(x)}=sinh(x) + tanh(x)[/tex]

Which makes it a little easier.

EDIT: The functions are not inverse trig functions they are hyperbolic functions incidentally.
 
Last edited:
Or, change variables: y = cosh(x), dy = sinh(x)dx.
 
Kurdt said:
Well the obvious chamge would be;

[tex]\frac{sinh(x)}{1+cosh(x)}=sinh(x) + tanh(x)[/tex]

Which makes it a little easier.

That equality isn't correct, although I'm pretty sure that that fraction works out to tanh(x/2) which might help a bit.
 
Do you see why

[tex]\int \frac{\sinh t}{1+\cosh t}{}dt =\ln\left(1+\cosh t\right) +\mathcal{C}[/tex]

The derivative of the denominator is the numerator.

Daniel.
 
d_leet said:
That equality isn't correct, although I'm pretty sure that that fraction works out to tanh(x/2) which might help a bit.

Thats a rather embarrassing schoolboy error that I wish I could blame on how late I was up last night, but its far too simple for that. :blushing:
 

Similar threads

How to Determine Correct Inverse Trig Angle?
  • · Replies 2 ·
Replies
2
Views
1K
Calculus 2 - Trig Integrals Question (Integrating cos^2x)
  • · Replies 3 ·
Replies
3
Views
2K
Using hyperbolic substitution to solve an integral
  • · Replies 18 ·
Replies
18
Views
4K
Integrating trig powers, why not by parts?
  • · Replies 5 ·
Replies
5
Views
2K
How to Prove the Integral of sin^7 x/(1+x^10) dx from -pi/2 to pi/2 is 0?
  • · Replies 14 ·
Replies
14
Views
2K
Trig Substitution Problem w/ tan substitution
  • · Replies 4 ·
Replies
4
Views
1K
Integration by u substitution for inverse trig formulas
  • · Replies 8 ·
Replies
8
Views
4K
Integration using inverse trig indentities?
  • · Replies 3 ·
Replies
3
Views
1K
Integration with Trig Sub: Simplifying the Square Root
  • · Replies 4 ·
Replies
4
Views
2K
How Do You Integrate (u^4 + 1) / (u^5 + 3u)?
  • · Replies 6 ·
Replies
6
Views
2K