Integration using cosh(t) or sinh(t)

In summary, for solving the equation \intdx/x squ rt of x^2 -1 using the appropriate substitution, it is recommended to first find a trig or hyper-trig identity that will simplify the integrand. Possible substitutions to consider are x=sec(t) for a trig substitution and x=cosh(t) for a hyperbolic function substitution. In this particular case, using x=sec(u) is the most efficient choice as it simplifies the integrand.
  • #1
Rubik
97
0
How do I go about solving the equation [tex]\int[/tex]dx/x squ rt of x^2 -1 using the appropriate substitution?
 
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  • #2
first find a trig or hyper-trig identity that will simplify the intergand

for example if you had
[tex] \int dx \sqrt{1-x^2} [/tex]

then
[tex] x = sin(x) [/tex]
[tex] x = cos(x) [/tex]

are both good choices as
[tex] sin^2(x) = 1 - cos^2(x) [/tex]
[tex] cos^2(x) = 1 - sin^2(x) [/tex]

in your case you will need find something that simplifies
[tex]sqrt{1-x^2} [/tex]
 
  • #3
Having [tex]\sqrt{x^2-1}[/tex] suggests a couple of possible substitutions:

[tex]\sec^2(\theta)-1=\tan^2(\theta),\ \ \text{so}\ \ x=\sec(t) [/tex] makes sense for a trig substitution.

[tex]\cosh^2(u)-1=\sinh^2(u),\ \ \text{so}\ \ x=\cosh(t) [/tex] makes sense for a hyperbolic function substitution.

Try either or both to see what works.
 
  • #4
Rubik said:
How do I go about solving the equation [tex]\int[/tex]dx/x squ rt of x^2 -1 using the appropriate substitution?

Just want to clarify, just what IS your integrand? It's very difficult to read. My best guess as to what it is is:

[tex]\int \frac{dx}{x \sqrt{x^2-1}}[/tex]

Is that correct?
 
  • #5
Yes that is it :)
 
  • #6
In that case I would definitely use the substitution x=sec(u). You'll find that everything cancels out very nicely.
 

1. What is the definition of cosh(t) and sinh(t)?

Cosh(t) and sinh(t) are both hyperbolic functions that are defined in terms of the exponential function e^x.

2. How do you use cosh(t) and sinh(t) in integration?

When integrating a function that involves cosh(t) or sinh(t), you can use the identities cosh(t) = (e^t + e^-t)/2 and sinh(t) = (e^t - e^-t)/2 to simplify the integration process.

3. Can cosh(t) and sinh(t) be used to solve differential equations?

Yes, cosh(t) and sinh(t) can be used to solve linear and non-linear differential equations involving exponential functions. They also have applications in physics, engineering, and other fields.

4. Are there any special properties of cosh(t) and sinh(t)?

Yes, cosh(t) and sinh(t) have several useful properties, such as being even and odd functions, respectively, and having a relationship to the pythagorean identity.

5. How do you graph a function involving cosh(t) or sinh(t)?

To graph a function involving cosh(t) or sinh(t), you can use a graphing calculator or plot points manually by substituting different values of t into the function and plotting the resulting points.

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