- #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

∫(x^2-1)^(-1/2) dx

and

∫x^(-1)*(1-x^2)^(-1/2) dx

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- Thread starter y_lindsay
- Start date

- #1

- 17

- 0

how to evaluate the indefinite integral

∫(x^2-1)^(-1/2) dx

and

∫x^(-1)*(1-x^2)^(-1/2) dx

∫(x^2-1)^(-1/2) dx

and

∫x^(-1)*(1-x^2)^(-1/2) dx

- #2

arildno

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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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@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 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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- #6

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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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