# Indefinite integral problem

1. Dec 25, 2007

### y_lindsay

how to evaluate the indefinite integral

∫(x^2-1)^(-1/2) dx

and

∫x^(-1)*(1-x^2)^(-1/2) dx

2. Dec 25, 2007

### arildno

Try to think like this:
Why are these integrals hard?

First hurdle:
There are ugly square roots here!

So, the next question is naturally:
What is the simplest way to "eliminate" a square root sign?
Ensuring that the radicand is a perfect square!
Then, the root and the square would eliminate each other.

Agreed?

Thus, tackle the first one:
If $$x^{2}-1=y^{2}$$
then $$x^{2}-y^{2}=1$$

Do you know of any functions that behaves like x and y here?

Possibly, you are familiar with the hyperbolic functions, Cosh(u) and Sinh(u) which satisfies for all u the identity: $$Cosh^{2}u-Sinh^{2}(u)=1$$

In terms of exponentials, we have the relations:
$$Cosh(u)=\frac{e^{u}+e^{-u}}{2},Sinh(u)=\frac{e^{u}-e^{-u}}{2}$$

Thus, we set $$x=Cosh(u)$$, whereby we transform the integral as follows:
$$\int\frac{dx}{\sqrt{x^{2}-1}}=\int\frac{1}{|Sinh(u)|}\frac{dx}{du}du$$

Now, we have that $$\frac{d}{du}Cosh(u)=Sinh(u)$$, whereby the integral reduces to:
$$sgn(Sinh(u))\int{du}=sgn(Sinh(u))u=|u|$$
since Sinh(u) and u has matching sign domains.

Okay so far?

3. Dec 25, 2007

### Defennder

Actually, arildno I don't think we need to introduce hyperbolic functions. They might work, but you're thinking too much. I could solve both of them by using normal (non-hyperbolic) trigo substitutions.

@OP:

You might find this trigo identity useful:

(tan θ)^2 = (sec θ)^2 - 1

Once you figured out the proper trigo subst for the first one from the clue above, you're done.

For the 2nd one, you need a different trigo substitution. The following identity is useful:

(sin θ)^2 = 1 - (cos θ)^2

4. Dec 25, 2007

### arildno

I LIKE hyperbolic functions, that's why I use them.

I am perfectly well aware of the sec substitution, but the result is more elegantly written in terms of hyperbolics/their inverses.
That's why I like them..

5. Dec 25, 2007

### HallsofIvy

Staff Emeritus
Defennnder, I also would tend to use trig functions, perhaps because that is what I first learned, but I noticed that arildno tends to use hyperbolic functions. It's certainly NOT a matter of "thinking too much"- one is as good as the other. One advantage I will conced to hyperbolic functions- you can use sinh and cosh rather than sec and tan which have more complicated integration and differentiation rules than sinh and cosh.

6. Dec 25, 2007

### y_lindsay

Thanks arildno, i haven't thought out the hyperbolic solution to the first problem.

yet to solve the 2nd problem,
is there any other way rather than using substitution x=sin u ?
I've thought out another possible substitution u=1/x, and then we could use the result of the firtst problem.
am i right?

7. Dec 25, 2007

### arildno

That should work nicely, yes.