hyperbolic functions in trig

[andyandy]

[latex] $$\int\nolimits \frac {x+1}{\sqrt{x^2-4}}dx $$


[/latex]


we have a formula in latex! And it only took 20 minutes



my question....

upon attempting to intergrate this, apparently one should realise that the substitution [latex]$x=2\cosh u$[/latex]
will allow an easy simplification.....

but what are the tell tale signs that a hyperbolic function could be useful for a substitution?

plus any interesting stuff about hyperbolic functions you wish to share/discuss.....

thanks

(ps the integral should have limits 2 and 3 if anyone can suggest how that should be done in latex....)

 

[andyandy]

is it just a case of looking for the form

[latex]
$$\int\nolimits \frac {a}{b}}dx $$[/latex]

where b is

[latex] \sqrt{cx^2 + d} [/latex]

??

Are there other cases where such a substitution is helpful?

 

[Unnamed]

but what are the tell tale signs that a hyperbolic function could be useful for a substitution?

Can't help you there...

(ps the integral should have limits 2 and 3 if anyone can suggest how that should be done in latex....)

But that I can do:
[latex] $$\int_{2}^{3} \frac {x+1}{\sqrt{x^2-4}}\, dx $$[/latex]

 

[andyandy]

Can't help you there...


But that I can do:
[latex] $$\int_{2}^{3} \frac {x+1}{\sqrt{x^2-4}}\, dx $$[/latex]

cheers - i was getting things like....

[latex] $$\int^\frac {x+1}{\sqrt{x^2-4}}\, dx $$[/latex]

 

[Yllanes]

but what are the tell tale signs that a hyperbolic function could be useful for a substitution?

plus any interesting stuff about hyperbolic functions you wish to share/discuss.....

thanks

(ps the integral should have limits 2 and 3 if anyone can suggest how that should be done in latex....)

When you have integrals of square roots of a cuadratic polynomial, you try to find a way to make them rational. There are some general rules. R denotes a rational function of x and the square root.
[latex]
\begin{align*}
\int R(x,\sqrt[n]{ax+b}) \qquad& \text{\sffamily is converted to a rational integral with $u=\sqrt[n]{ax+b}$}\\
\int R(x,\sqrt{a^2-x^2})\qquad & \text{\sffamily to rational with $x=a\sin u$}\\
\int R(x,\sqrt{x^2+a}) \qquad &\text{\sffamily to rational with $u=x+\sqrt{x^2+a}$}
\end{align*}
[/latex]

It is easy to prove this. For example, for the last one [latex]
$(u-x)^2=x^2+a=u^2+x^2-2xu\quad\Longrightarrow\quad x = \frac{u}{2} -\frac{a}{2u}\quad\Longrightarrow\quad \mathrm{d}x= \frac12 + \frac a{2u^2}$
[/latex]. Why wouldn't u = sqrt{x^2+a} work?

Your example is similar to the second one, but with a different sign. Remember that cosh^2 x - sinh^2 x = 1, so the hyperbolic functions take care of this changed sign. In your example \sqrt{x^2-4} = 2\sqrt{cosh^2 u -1} = 2 sinh u. At the same time, dx= 2 senh u du and. So doing this takes care of the square root, the tricky part. After that, the integration is immediate.

The idea is always to get rid of the roots. Rational integrals of trigonometric functions are easily dealt with (they may be long, but they are systematic).

 

[SirPhilip]

Ha. My function is already hyperbole.

 

[Schneibster]

What, ANYTHING?

OK, how's this grab you:

Turns out that hyperbolic trig functions map 1:1 to the equations needed to describe mechanics in special relativity. For example, the equation for a Lorentz boost turns out to be:
t' = t
x' = (cosh s)x + (sinh s)y
y' = (sinh s)x + (cosh s)y
z' = z

Is there a reason for this? Yes! It turns out that the geometrical arrangement of time with respect to space is hyperbolic, not circular the way the three spatial dimensions are geometrically arranged with respect to one another. Space, then, is curved into a hyperboloid with respect to time, and the time axis forms the axis of symmetry of the hyperboloid (assuming we're talking about flat spacetime). These less-familiar forms of the Lorentz transform equations are nearly identical to the equations, using ordinary trig functions, for rotations in space. In fact, physicists often talk about the "rapidity" of a moving frame (or object), which is a way of thinking about the hyperbolic angle of rotation that the frame or object has as a result of its velocity.

 

[andyandy]

When you have integrals of square roots of a cuadratic polynomial, you try to find a way to make them rational. There are some general rules. R denotes a rational function of x and the square root.
[latex]
\int R(x,\sqrt[n]{ax+b}) \qquad& \text{\sffamily is converted to a rational integral with $u=\sqrt[n]{ax+b}$}\ [/latex]

working this one through, the formula for the finished integral is (I think)

[latex]
$$\frac{a}{ndu}\ln(u^{n-1}) [/latex]

with absolutes rather than brakets and a C knocking around....

hang on....not sure if that's right...

 

[andyandy]

disregard that last post....it's utter nonsense

 

[Yllanes]

working this one through, the formula for the finished integral is (I think)

[latex]
$$\frac{a}{ndu}\ln(u^{n-1}) [/latex]

with absolutes rather than brakets and a C knocking around....

hang on....not sure if that's right...

The du can't be in the denominator. I'll give you an example of these integrals:[latex]$\int x (1+x)^{1/4}$[/latex]. You set u = (1+x)^(1/4), so that x = u^4 -1 and dx = 4u^3 du. So now we have

[latex]
$$
\int x (1+x)^{1/4}\ \mathrm{d}x = \int 4 (u^8-u^4)\ \mathrm{d}u = \frac{4u^9}{9} - \frac{4u^5}{5} = \frac49 (1+x)^{9/4} - \frac45 (1+x)^{5/4}
$$
[/latex]

Your best bet is to pick a book with many proposed integrals (even if they are not solved, Spivak's or Apostol's Calculus are good for this) and practice until you see the change of variable. You need a book, because many times a very simple integrand has no primitive (without recurring to special functions) or requires very sophisticated methods to tackle. You can then check your answers with Wolfram's integrator (or differentiate and compare with the integrand). If you have access to Maple, it has a very good tutor for this kind of integrals. You can write one and ask for hints step by step.

 

[andyandy]

Your best bet is to pick a book with many proposed integrals (even if they are not solved, Spivak's or Apostol's Calculus are good for this) and practice until you see the change of variable. You need a book, because many times a very simple integrand has no primitive (without recurring to special functions) or requires very sophisticated methods to tackle. You can then check your answers with Wolfram's integrator (or differentiate and compare with the integrand). If you have access to Maple, it has a very good tutor for this kind of integrals. You can write one and ask for hints step by step.

yes....i said to disregard that post

I've studied this before - but it's been a few years - and i'm trying to get back to speed - and realising i've forgotten rather a lot
thanks for the wolfram link - should help quite a bit....