# Using hyperbolic substitution to solve an integral

• Howard Fox
In summary: It was dx but then we substituted it to du, no?In summary, the student is having trouble simplifying the denominator of an integral.
Howard Fox

## Homework Equations

So the question is asking to solve an integral and to use the answer of that integral to find an additional integral. With part a, I don't have much problem, but then I don't know how to apply the answer from it to part b. I know I should subsitute all the x's in part b with a trig function, right? But which one? [/B]

## The Attempt at a Solution

This is my solution for part a, I am not sure how to work on part b
[/B]

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You are replacing ##\sinh^2x## with ##1+\cosh^2x##. Check your algebra.

Also, the derivative of cosh(x) is sinh(x), not -sinh(x).

Howard Fox
Okay, I tried to correct the previous mistakes,

Hope this one is good? Thank you

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It is good. Look at part (a) and then look at part (b). What kind of hyperbolic substitution is worth trying in (b)?

kuruman said:
It is good. Look at part (a) and then look at part (b). What kind of hyperbolic substitution is worth trying in (b)?
I really don't know. I have trouble reconciling the two parts. I guess we could substitute x with asinh?

Howard Fox said:
I really don't know. I have trouble reconciling the two parts. I guess we could substitute x with asinh?
Excellent idea! What do you think ##a## ought to be? Hint: Look at the denominator.

kuruman said:
Excellent idea! What do you think ##a## ought to be? Hint: Look at the denominator.
It should be 3?

kuruman said:
Excellent idea! What do you think ##a## ought to be? Hint: Look at the denominator.

So the final integral should look something like this?

This does seem very complicated to solve though!

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Howard Fox said:
It should be 3?
Yes.
Howard Fox said:
So the final integral should look something like this?
View attachment 239808
This does seem very complicated to solve though!
That's because your algebra has been careless (again). Take the time to do it right. Then you will see that
(a) the denominator simplifies.
(b) you do not have a proper integral because you substituted ##1## for ##dx##.
(c) the numerator is incorrect.

kuruman said:
Yes.

That's because your algebra has been careless (again). Take the time to do it right. Then you will see that
(a) the denominator simplifies.
(b) you do not have a proper integral because you substituted ##1## for ##dx##.
(c) the numerator is incorrect.
Uhm, is this any better?

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Howard Fox said:
Uhm, is this any better?

View attachment 239812
Much better. Now concentrate on simplifying the denominator. How could you get rid of the radical? How does one usually get rid of a radical?

kuruman said:
Much better. Now concentrate on simplifying the denominator. How could you get rid of the radical? How does one usually get rid of a radical?
Raising both denominator and numerator to the power of two?

You would do that if the expression under the radical is not a perfect square. Certainly the ##9## in the denominator is a perfect square. What about the rest? Hint: Why did you choose ##3## for the substitution ##x=3\sinh v##?

kuruman said:
You would do that if the expression under the radical is not a perfect square. Certainly the ##9## in the denominator is a perfect square. What about the rest? Hint: Why did you choose ##3## for the substitution ##x=3\sinh v##?
/

Just take the radical of the values like this?

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Howard Fox said:
/View attachment 239813
Just take the radical of the values like this?
Absolutely not ! Is ##\sqrt{2^2+1^2}=2+1=3~##?

kuruman said:
Absolutely not ! Is ##\sqrt{2^2+1^2}=2+1=3~##?
Okay, went back to my textbook. This should be the way to simplify the denominator, right?

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Right, except there is more carelessness. What is your integration variable?

kuruman said:
Right, except there is more carelessness. What is your integration variable?
It was dx but then we substituted it to du, no?

## 1. What is hyperbolic substitution?

Hyperbolic substitution is a mathematical technique used to simplify and solve integrals involving hyperbolic functions such as sine, cosine, and tangent.

## 2. When should I use hyperbolic substitution?

Hyperbolic substitution is most useful when dealing with integrals that contain both algebraic and hyperbolic terms.

## 3. How do I use hyperbolic substitution to solve an integral?

To use hyperbolic substitution, you must first identify which hyperbolic function to substitute in for the variable. Then, use the corresponding trigonometric identity to rewrite the integral in terms of the hyperbolic function. Finally, solve the integral using standard integration techniques.

## 4. What are the benefits of using hyperbolic substitution?

Hyperbolic substitution can simplify complex integrals and make them easier to solve. It can also help to identify patterns and relationships between different integrals.

## 5. Are there any limitations to using hyperbolic substitution?

Hyperbolic substitution is only useful for integrals involving hyperbolic functions. It may not be applicable or helpful for other types of integrals. Additionally, it may not always lead to a closed-form solution and numerical methods may be required to solve the integral.

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