# Integral involving hyperbolic functions

• DryRun
In summary, the integral of x over the square root of a quadratic expression can be simplified by completing the square and using substitution. The final answer can be found by evaluating the integrated expression with the given limits. Additionally, the original integrand can be simplified by rewriting the numerator as x - 1/2 + 1/2. Using these techniques, the integral can be solved and the final answer can be obtained.
DryRun
Gold Member
Homework Statement
Find $$\int \frac{x}{\sqrt{2x^2-2x+1}}\,dx$$

The attempt at a solution

First, i complete the square for the quadratic expression:
$$2x^2-2x+1=2((x-\frac{1}{2})^2+\frac{1}{4})$$
$$\int \frac{x}{\sqrt{2x^2-2x+1}}\,dx=\int \frac{x}{\sqrt 2 \sqrt{(x-\frac{1}{2})^2+\frac{1}{4}}}\,dx$$

Using substitution: Let $(x-\frac{1}{2})=\frac{1}{2}\sinh u$

I eventually get this expression, after integrating:
$$\frac{1}{2\sqrt 2}(\cosh (\sinh^{-1}(2x-1)+\sinh^{-1}(2x-1))$$

But i don't know how to simplify this any further, without getting a complicated answer. The final answer is: $\frac{\sinh^{-1}1}{\sqrt 2}$

Last edited:
hi sharks!

(you haven't said what the limits are)
sharks said:
I eventually get this expression, after integrating:
$$\frac{1}{2\sqrt 2}(\cosh (\sinh^{-1}(2x-1)+\sinh^{-1}(2x-1))$$

But i don't know how to simplify this any further

cosh(sinh-1x) = √(x2 + 1)

(btw, you could have simplified the original integrand by rewriting the numerator as x - 1/2 + 1/2 )

Hi tiny-tim!

Congratulations on your PF award!

Doh! I forgot to include the limits.

$$\int^1_0 \frac{x}{\sqrt{2x^2-2x+1}}\,dx$$

If i evaluate over the limits:
$$\frac{1}{2\sqrt 2}(\cosh (\sinh^{-1}(2x-1)+\sinh^{-1}(2x-1))$$
Indeed, i get the answer!

I am trying to understand your suggestion:
$$\cosh (\sinh^{-1}x) = \sqrt{(x^2 + 1)}$$
It's too complicated to prove? I can't get the R.H.S. from the L.H.S.

Here is where I've reached...
$$\frac{x^2+x\sqrt{x^2+1}+1}{x+\sqrt{x^2+1}}$$

Last edited:
sharks said:
I am trying to understand your suggestion:
$$\cosh (\sinh^{-1}x) = \sqrt{(x^2 + 1)}$$

if y = sinh-1x

then x = sinhy,

so √(x2 + 1) = √(sinh2y + 1) = coshy = cosh(sinh-1x)

(if you think about it, this is obvious! )

I have also tried to solve this problem using your other suggestion - rewriting the numerator as x - 1/2 + 1/2, and i was able to get the answer.

Thank you for your insightful suggestions, tiny-tim.

## 1. What are hyperbolic functions?

Hyperbolic functions are mathematical functions that are analogous to trigonometric functions. They are defined in terms of the exponential function, and are useful in solving problems involving hyperbolas, hence the name "hyperbolic". The most commonly used hyperbolic functions are sinh, cosh, and tanh.

## 2. How are hyperbolic functions related to integrals?

Hyperbolic functions can be used to solve integrals involving hyperbolic expressions. These integrals often arise in physics and engineering problems involving exponential growth or decay, and can be solved using various techniques such as substitution or integration by parts.

## 3. What is the general form of an integral involving hyperbolic functions?

The general form of an integral involving hyperbolic functions is ∫f(x)g(x)dx, where f(x) and g(x) are hyperbolic functions such as sinh, cosh, or tanh. This form is similar to integrals involving trigonometric functions, but the techniques used to solve them may differ.

## 4. What is the most common technique used to solve integrals involving hyperbolic functions?

The most common technique used to solve integrals involving hyperbolic functions is substitution. This involves substituting a variable with a hyperbolic expression, and then using algebraic manipulation and integration rules to solve the resulting integral.

## 5. Are there any special cases or identities for integrals involving hyperbolic functions?

Yes, there are several special cases and identities for integrals involving hyperbolic functions. For example, some integrals can be solved using the hyperbolic Pythagorean identity, while others can be transformed into integrals involving trigonometric functions. It is important to be familiar with these identities in order to solve integrals involving hyperbolic functions efficiently.

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