Indefinite integral problem

  • Thread starter y_lindsay
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  • #1
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how to evaluate the indefinite integral

∫(x^2-1)^(-1/2) dx

and

∫x^(-1)*(1-x^2)^(-1/2) dx
 

Answers and Replies

  • #2
arildno
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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?
Answer:
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 [tex]x^{2}-1=y^{2}[/tex]
then [tex]x^{2}-y^{2}=1[/tex]

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: [tex]Cosh^{2}u-Sinh^{2}(u)=1[/tex]

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

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

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

Okay so far?
 
  • #3
Defennder
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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
arildno
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I LIKE hyperbolic functions, that's why I use them. :smile:

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
HallsofIvy
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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
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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
arildno
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That should work nicely, yes.
 

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