- #1

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- 316

- Homework Statement
- $$\int \sqrt{4+x^2}

dx$$

- Relevant Equations
- hyperbolic equations

still typing...checking latex

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correct! i just checked...i ended up with,$$\int \sqrt{(8-4\sinh^2 u)}⋅ 2 \sinh u du$$...not looking good!

- #1

- 2,444

- 316

- Homework Statement
- $$\int \sqrt{4+x^2}

dx$$

- Relevant Equations
- hyperbolic equations

still typing...checking latex

Physics news on Phys.org

- #2

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- 316

This is a textbook question...the steps to solution are pretty easy...they indicated use of;

##x=2\sinh u## that will realize

$$\int \sqrt{4+x^2} dx=\int 2\cosh u⋅ 2\cosh u du=4\cosh^2 u du$$

...from here the steps are pretty clear up to the required solution.

Ok to my question now; Could we also use;

##x=2\cosh u## instead of ##x=2\sinh u##?

Cheers

##x=2\sinh u## that will realize

$$\int \sqrt{4+x^2} dx=\int 2\cosh u⋅ 2\cosh u du=4\cosh^2 u du$$

...from here the steps are pretty clear up to the required solution.

Ok to my question now; Could we also use;

##x=2\cosh u## instead of ##x=2\sinh u##?

Cheers

Last edited:

- #3

- 25,774

- 17,346

That doesn't look so promising to me.chwala said:Ok to my question now; Could we also use;

##x=2\cosh u## instead of ##x=2\sinh u##?

Cheers

- #4

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- 316

I thought it will be other way round...let me try and see what comes out of it...will share later. Cheers mate.PeroK said:That doesn't look so promising to me.

- #5

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- 11,016

No. The point is to rewrite ##\sqrt{4 + x^2} = \sqrt{4(1+\sinh^2(u))} = 2 \cosh(u)## using the hyperbolic one. If you would have attempted to use ##x = 2 \cosh(u)## instead you would have ended up with ##\sqrt{4 + x^2} = 2\sqrt{1+\cosh^2(u)}## and ##1 + \cosh^2(u)## does not have any particular simplification in terms of the hyperbolic one. If you would attempt to replace ##\cosh^2(u)## by ##\sinh^2(u)## using the hyperbolic one you would instead end up with ##2 + \sinh^2(u)##, which doesn't make you any happier.chwala said:Ok to my question now; Could we also use;

##x=2\cosh u## instead of ##x=2\sinh u##?

- #6

Mentor

- 37,349

- 9,533

With this substitution, the integral ##\int \sqrt{x^2 + 4}~dx## becomes ##4\int \sec^3(\theta)~d\theta##, an integral so well-known there's a wikipedia article devoted to it.

- #7

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Thanks @Mark44 Let me study on this approach...Mark44 said:

With this substitution, the integral ##\int \sqrt{x^2 + 4}~dx## becomes ##4\int \sec^3(\theta)~d\theta##, an integral so well-known there's a wikipedia article devoted to it.

- #8

Mentor

- 37,349

- 9,533

The picture I was describing looks like this. The hypotenuse is ##\sqrt{x^2 + 4}##.

- #9

Science Advisor

Gold Member

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Its a good rule of thumb to consider trigonometric substitutions when seeing the likes of ## \sqrt {a^2 \pm x^2}dx ##

When it's a +, consider secant, tangent;

when it's a - , consider sin, cos

Subbing ##u=ax## in each case, as

##1 \pm x^2## is " translatsble" to either of these.

Edit: Of course, consider the issue of limits of integration.

When it's a +, consider secant, tangent;

when it's a - , consider sin, cos

Subbing ##u=ax## in each case, as

##1 \pm x^2## is " translatsble" to either of these.

Edit: Of course, consider the issue of limits of integration.

Last edited:

- #10

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correct! i just checked...i ended up with,Orodruin said:No. The point is to rewrite ##\sqrt{4 + x^2} = \sqrt{4(1+\sinh^2(u))} = 2 \cosh(u)## using the hyperbolic one. If you would have attempted to use ##x = 2 \cosh(u)## instead you would have ended up with ##\sqrt{4 + x^2} = 2\sqrt{1+\cosh^2(u)}## and ##1 + \cosh^2(u)## does not have any particular simplification in terms of the hyperbolic one. If you would attempt to replace ##\cosh^2(u)## by ##\sinh^2(u)## using the hyperbolic one you would instead end up with ##2 + \sinh^2(u)##, which doesn't make you any happier.

$$\int \sqrt{(8-4\sinh^2 u)}⋅ 2 \sinh u du$$...

not looking good!

Last edited:

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