# Integral involving hyperbolic functions

1. Apr 15, 2012

### sharks

The problem statement, all variables and given/known data
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: Apr 15, 2012
2. Apr 15, 2012

### tiny-tim

hi sharks!

(you haven't said what the limits are)
cosh(sinh-1x) = √(x2 + 1)

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

3. Apr 15, 2012

### sharks

Hi tiny-tim!

Doh! I forgot to include the limits.

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

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

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: Apr 15, 2012
4. Apr 15, 2012

### tiny-tim

if y = sinh-1x

then x = sinhy,

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

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

5. Apr 15, 2012

### sharks

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.