Integration and inverse trig functions

In summary, the student is struggling with integrating trig functions and specifically with the integral of sinh(x)/1+cosh(x). They are unsure of how to start and are wondering if an identity can be used. Another student suggests using the identity y = cosh(x), dy = sinh(x)dx to simplify the problem. Another student points out that the original equality given by the first student is incorrect, but it can be simplified to tanh(x/2). Finally, another student suggests using the equation \int \frac{\sinh t}{1+\cosh t}{}dt =\ln\left(1+\cosh t\right) +\mathcal{C} which can be derived from the derivative of the denominator
  • #1
mavsqueen06
7
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


?
 
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  • #2
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:
  • #3
Or, change variables: y = cosh(x), dy = sinh(x)dx.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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:
 

1. What is integration?

Integration is a mathematical process that involves finding the area under a curve on a graph. It is the reverse process of differentiation and is used to solve problems in many fields such as physics, engineering, and economics.

2. What are inverse trig functions?

Inverse trig functions are mathematical functions that undo the effects of their corresponding trigonometric functions. They are used to find the angle or side length of a right triangle when given other known values.

3. Why do we use integration in calculus?

Integration is an important tool in calculus because it allows us to solve problems involving rates of change, accumulation, and optimization. It is also used to find the area and volume of irregular shapes and objects.

4. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, meaning that the integration is performed over a specific interval. An indefinite integral, on the other hand, does not have limits and represents a family of functions that differ only by a constant.

5. How do we integrate inverse trig functions?

The process of integrating inverse trig functions involves using trigonometric identities and substitution techniques. It is important to also have a good understanding of derivatives and the fundamental theorem of calculus in order to successfully integrate these functions.

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