# Hyperbolic functions/integration

1. Dec 11, 2007

### Zoe-b

1. The problem statement, all variables and given/known data
Show that: (I don't have any way of showing it properly so have written it in words)

'The integral from 5 to 3 of (sq root (x^2 - 9) ) with respect to x' = 10 - (9/2) * ln 3

I hope that makes sense!

2. Relevant equations
Well I'm not sure.. I *think* that the fact the integral of 1/(sq root (x^2 - a^2)) is arcosh (x/a) is relevant.

3. The attempt at a solution
This is part of the 'hyperbolic functions' section on my exam; so the first thing I know is that they will come into it somehow. I cannot solve this by inspection, or by simply comparing it to all the formulae on my sheet.. I also tried to solve by parts (using difference of two squares to give me a root multiplied by another root) but found that this still gave me something I do not know how to integrate. I know if it was 1/(sq root (x^2 - 9)) it would be easy; but cannot work out how the integral of a particular thing and the integral of 1/it are related.

2. Dec 11, 2007

### HallsofIvy

Staff Emeritus
It is true that $cosh^2(\theta)- sinh^2(\theta)= 1$ so, dividing both sides by $cosh^(\theta)$, $1- tanh^2(\theta)= csch^2(\theta)$ . That is, if you let $x= 3coth(\theta), [itex]9- x^2= 9- 9coth^2(\theta)= 9csch^2(\theta)$ so $\sqrt{9- x^2}= \sqrt{9csch^2(\theta)}= 3csch(\theta)$. Of course, $3(coth(\theta))'= 3(1- coth(\theta))$ so $dx= 3(1- coth(\theta))d\theta$.
The integral becomes
$$9\int coth(\theta)(1- coth(\theta)d\theta$$

3. Dec 11, 2007

### malty

Don't think arccos is relevant. . .

Hmm I don't think a hyperbolic substition is necessary here . ..need to check my calculations ... .

//Argh forgot to square the minus// Nevermind:

Last edited: Dec 11, 2007
4. Dec 11, 2007

### Zoe-b

Right. eek.. I really didn't follow that! sorry. right..

what does [itex] mean?? and how did you get from sq root(x^2 - 9) to sq root (9 - x^2)? without getting into complex numbers..

5. Dec 11, 2007

### malty

I dunno how he did that, because that's what's worrying me each time I do this integral I get a complex function ... *not having a good day:(*

[ ite k or latek" is just a thing for typing lovely equations, e.g $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Last edited: Dec 11, 2007
6. Dec 11, 2007

### Zoe-b

Well I tried using the substitution x = 3coth theta... though I'll call theta y to make it easier to type.
I then get x^2 - 9 = 9coth^2 y - 9
root (x^2 - 9) = 3(cosech y)

therefore the original question is:
the integral from x=5 to x=3 of 1/(3 (cosech y)) dx.

dx/dy = -3(cosech y) ... I think, not so sure on my maths there. now I get stuck again as this seems to cancel to give no variable to integrate..

7. Dec 11, 2007

### malty

...oopps... never mind that

$$\frac{dx}{d\theta}=3 cosch^2 ({\theta})$$ .. .

Starting over grrr

Yay i've finally got it!! ... I think

Last edited: Dec 11, 2007
8. Dec 11, 2007

### Zoe-b

I found a solution online (the cheats way out I know!).. but I'll do it properly when I get time. Its easier to use a different substitution (x = 3 cosh y, if I remember rightly).
Thanks anyway.

9. Jan 19, 2012

### someriraq

sorry HallsofIvy : 1- tanh^2 (theta)= sech^2(theta) !!
thanks