MHB Integration with trig and hyperbolic substitutions

[Dethrone]

Okay, okay, I'm making my head work a little bit.
Let's take a look at how we can define $\cosh$ and $\sinh$ geometrically. (Worried)
So if we have an $x$ with $x=\cosh \alpha$, it follows that double the red area, which is $\alpha$, is equal to $\arcosh x$.
Taking the $\sinh$ from that, we get the y-coordinate of the corresponding right triangle.

Since we have $x^2 - y^2 = 1$, it follows that:
$$y^2 = x^2 - 1$$
$$y = \sinh(\arcosh x) = \sqrt{x^2 - 1}$$
(Mmm)

I'm trying to understand this.
Say we have to find $\cosh(\arsinh(x))$.
$u=\arsinh(x)$ and $x^2-y^2=1$, where $\cosh (u)=x$
$x=\cosh(\arsinh(x))=\sqrt{1+y^2}$

Why is my answer in terms of $y$? Did I make a mistake?

 

[I like Serena]

I've actually memorized all those too...but I was referring to a more challenging integral:

$$\displaystyle \int \frac{u^2du}{\left(u^2-a^2\right)^{3/2}}$$

Circular or hyperbolic?

Looks hyperbolic doesn't it? (Thinking)

Circular: $y=\sqrt{1-x^2} \Rightarrow x^2+y^2=1$.
Hyperbolic: $y=\sqrt{x^2-1} \Rightarrow x^2-y^2=1$.

 

[I like Serena]

I'm trying to understand this.
Say we have to find $\cosh(\arsinh(x))$.
$u=\arsinh(x)$ and $x^2-y^2=1$, where $\cosh (u)=x$
$x=\cosh(\arsinh(x))=\sqrt{1+y^2}$

Why is my answer in terms of $y$? Did I make a mistake?

That's quite alright. (Smile)

(Although it should be $x=\cosh(\arsinh(y))=\sqrt{1+y^2}$. )[/color]

Often we write y as function of x, but there is nothing wrong with writing x as a function of y.
This is similar to determining an inverse. (Nerd)

 

[Dethrone]

How far do you get with the antiderivatives of:
$$\int \frac {dx} {1+x^2} \tag{1}$$
$$\int \frac {dx} {1-x^2} \tag{2}$$
$$\int \frac {dx} {\sqrt{1 + x^2}} \tag{3}$$
$$\int \frac {dx} {\sqrt{1- x^2}} \tag{4}$$
$$\int \frac {dx} {\sqrt{x^2-1}} \tag{5}$$
$$\int \frac {x\, dx} {\sqrt{1 + x^2}} \tag{6}$$
(Wondering)

1) is the inverse tangent, 2) is the inverse hyperbolic tangent, 3) is the inverse hyperbolic sine, 4) is the inverse hyperbolic cosine, 5) is the inverse sine, 6) I don't know, but it requires u-sub because I see it's derivative in the numerator. This is how I would solve these...but I don't see these are related to my other question?

 

[I like Serena]

1) is the inverse tangent, 2) is the inverse hyperbolic tangent, 3) is the inverse hyperbolic sine, 4) is the inverse hyperbolic cosine, 5) is the inverse sine,

There's a couple of mistakes in there.

6) I don't know, but it requires u-sub because I see it's derivative in the numerator.

As a challenge, can you do it without an u-sub? (Wondering)

This is how I would solve these...but I don't see these are related to my other question?

That's what you get from mutually posting quicker than being able to respond. (Rofl)

 

[Dethrone]

That's quite alright. (Smile)

(Although it should be $x=\cosh(\arsinh(y))=\sqrt{1+y^2}$. )[/color]

Often we write y as function of x, but there is nothing wrong with writing x as a function of y.
This is similar to determining an inverse. (Nerd)

I want to write it as a function of $x$, let me try again...
$y=\sinh(x)$ and $x=\arsinh(y)$
Determine $\cosh(\arsinh(y))$. We know $\cosh^2(x)-\sinh^2(x)=1$
$\cosh(x)=\sqrt{1+y^2}$?

 

[Dethrone]

There's a couple of mistakes in there.
As a challenge, can you do it without an u-sub? (Wondering)
That's what you get from mutually posting quicker than being able to respond. (Rofl)

I knew I probably made some mistakes there because they were all so similar, I kind of got tangled up.
I don't know...partial integration?
$$\int x \,d(\arsinh(x))$$

 

[I like Serena]

I want to write it as a function of $x$, let me try again...
$y=\sinh(x)$ and $x=\arsinh(y)$
Determine $\cosh(\arsinh(y))$. We know $\cosh^2(x)-\sinh^2(x)=1$
$\cosh(x)=\sqrt{1+y^2}$?

You're mucking it up! (Doh)

It should be (aligned with the graph):
$y=\sinh(\alpha)$ and $\alpha=\arsinh(y)$

Determine $\cosh(\arsinh(y))$. We know $x^2-y^2=1$
$x = \cosh(\alpha)=\sqrt{1+y^2}$

To write it as function of $x$, we need to swap $x$ and $y$ to get:
$\cosh(\arsinh(x)) = \sqrt{1+x^2}$ (Whew)

 

[I like Serena]

I knew I probably made some mistakes there because they were all so similar, I kind of got tangled up.
I don't know...partial integration?
$$\int x \,d(\arsinh(x))$$

Note that the numerator is more or less the derivative of the argument of the square root in $$\int \frac{x\,dx}{\sqrt{1+x^2}}$$.
Since the derivative of $\sqrt u$ is $\frac{1}{2\sqrt{u}}$ (did you know that by heart?), care to reconsider? (Thinking)

 

[Dethrone]

Yes...it's simply $\int \,d(\sqrt{1+x^2})$, but that takes a lot of brain power to work through (Muscle), and it's just a u-substitution in disguise.

How will this help with more complicated integrals?

- - - Updated - - -

Looks hyperbolic doesn't it? (Thinking)

Circular: $y=\sqrt{1-x^2} \Rightarrow x^2+y^2=1$.
Hyperbolic: $y=\sqrt{x^2-1} \Rightarrow x^2-y^2=1$.

I'm pretty sure it looks circular too, with the relationship $1+\tan^2(x)=\sec^2(x)$

 

[I like Serena]

Yes...it's simply $\int \,d(\sqrt{1+x^2})$, but that takes a lot of brain power to work through (Muscle), and it's just a u-substitution in disguise.

How will this help with more complicated integrals?

It just helps to know and recognize those "standard" integrals.
When you get a more complicated integral, the challenge is to reduce it to one of the standard integrals.

 

[Dethrone]

So how would you evaluate this by recognition of standard integrals? (Wondering)

$$\displaystyle \int \frac{x^2dx}{\left(x^2-9\right)^{3/2}}$$

 

[Dethrone]

When $0<\theta<\frac{\pi}{2}$, (1st quadrant) then
$$= \int \frac{\sec\theta \tan\theta\,d\theta}{\tan\theta}=\int \sec\left({\theta}\right) \,d \theta$$
When $-\frac{\pi}{2}<\theta<\pi$, (2nd quadrant) then
$$= -\int \frac{\sec\theta \tan\theta\,d\theta}{\tan\theta}=-\int \sec\left({\theta}\right) \,d \theta$$

Correct :D?

I went back and rethought this. If we were to evaluate both cases, we have:

When $0<\theta < \pi/2$:
$=\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$
When $\pi/2 < \theta < \pi$:
$=-\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$

Now this answer is different from our hyperbolic substitution. The hyperbolic substitution only gave us $=\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$, not the negative version too...

Spoiler

Just to make this thread hard-to-follow (Giggle)

 

[I like Serena]

So how would you evaluate this by recognition of standard integrals? (Wondering)

$$\displaystyle \int \frac{x^2dx}{\left(x^2-9\right)^{3/2}}$$

Well... I guess you would have a clue about $$\int \frac{dx}{\left(x^2-9\right)^{1/2}}$$, right?

It's just that the power of the denominator is 1 point too high.
But we can get that by taking a derivative.

Let's see...

$$d\left(\frac{1}{\left(x^2-9\right)^{1/2}}\right)
= -\frac 12 \cdot \frac{1}{\left(x^2-9\right)^{3/2}} \cdot 2x\,dx
= \frac{-x\,dx}{\left(x^2-9\right)^{3/2}}$$

Hmm...
Perhaps we can use that in a partial derivative... (Thinking)

$$\int \frac{x^2dx}{\left(x^2-9\right)^{3/2}}
=\int -x\,d\left(\frac{1}{\left(x^2-9\right)^{1/2}}\right)
= ...
$$

Where would that go? (Wondering)

 

[I like Serena]

I went back and rethought this. If we were to evaluate both cases, we have:

When $0<\theta < \pi/2$:
$=\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$
When $\pi/2 < \theta < \pi$:
$=-\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$

Now this answer is different from our hyperbolic substitution. The hyperbolic substitution only gave us $=\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$, not the negative version too...

I think that should come out as something like:

When $\pi/2 < \theta < \pi$:
$=-\ln \left| -x+\sqrt{x^2-a^2} \right|+K, \text{ where K=+$\ln\left({a}\right) + C$}$

(Thinking)

 

[Dethrone]

Holy...that's next level integration. In school and in most textbooks, we were taught to use trig substitutions, and in this case, $x=3 \sec (\theta)$. It seems like you don't even use it...

We get:
$$=\frac{-x}{\sqrt{x^2-9}}+\int \frac{dx}{\sqrt{x^2-9}}+C$$
$$=\frac{-x}{\sqrt{x^2-9}}+ \arcosh(x/3)+C$$

I think that should come out as something like:

When $\pi/2 < \theta < \pi$:
$=-\ln \left| -x+\sqrt{x^2-a^2} \right|+K, \text{ where K=+$\ln\left({a}\right) + C$}$

(Thinking)

But doesn't that give us two different answers? We only got one from the hyperbolic method.

 

[I like Serena]

Holy...that's next level integration. In school and in most textbooks, we were taught to use trig substitutions, and in this case, $x=3 \sec (\theta)$. It seems like you don't even use it...

We get:
$$=\frac{-x}{\sqrt{x^2-9}}+\int \frac{dx}{\sqrt{x^2-9}}+C$$
$$=\frac{-x}{\sqrt{x^2-9}}+ \arcosh(x/3)+C$$

Whot? I was only trying to recognize the standard integrals as you suggested. (Crying)

But yeah, that's it! (Yes)

To be honest, I don't really like those complicated trig substitutions. (Bandit)

But doesn't that give us two different answers? We only got one from the hyperbolic method.

Then I suspect there must be a mistake with a minus sign somewhere.

 

[Dethrone]

This is crazy..mind blown...guess I don't need trig substitutions anymore (Giggle) I guess I'll practice integration by recognition, just as the pros do it. (Cool)

Spoiler

This is the method you use to integrate these functions, right? If so, then I should practice it...

$$=-\int \sec(\theta) \,d\theta +C$$

We are considering when $x<-a$, with the substitution $x=a \sec(\theta)$. Solving the inequality $a \sec(\theta)<-a$, we get $\sec (\theta)<-1$, which occurs when $\pi/2 < \theta <\pi$.

$$=-\ln\left| \sec(\theta)+\tan(\theta) \right|+C$$

From a right triangle, we obtain the value of $\sec(\theta)=x/a$ and $\tan(\theta)=\sqrt{x^2-a^2}/a$, which, in the 2nd quadrant, secant is negative.

$$=-\ln\left| -x+\sqrt{x^2-a^2}\right|+C$$
$$=-\ln\left| x-\sqrt{x^2-a^2}\right|+C$$
$$=\left| x+ \sqrt{x^2-a^2}\right|+C$$

Is that correct? This means that you can neglect the absolute value from the secant substitution because both answers when $x>a$ and $x<-a$ are equivalent!

People! Don't convert my thread into a spam thread, please (Crying)

Also, do all even trig function have two inverses? $y=\cos(\theta)=\cos(-\theta)$, therefore, $\arccos(y)$ and $-\arccos(y)$. (Whew)

 

[Dethrone]

Let's distinguish cases.
If $\theta>0$, then we have $dx = a \sinh(\theta)\,d\theta = a \text{ sgn}(\theta)\sinh(\theta)\,d\theta$.
And if $\theta<0$, we have $dx = -a \sinh(\theta)\,d\theta = a \text{ sgn}(\theta)\sinh(\theta)\,d\theta$.
There you go. (Mmm)

I think I finally understand this, but when I read back, it's making less sense. Why are we saying $\theta <0$ and $\theta >0$? Should it not be $x>a$ and $x<-a$?

Let's look at the graph:

Essentially, we want the two answers $\arcosh(x/a)$ and $-\arcosh(-x/a)$, which are defined when $x\ge a$ and $x\le -a$ respectively.

This is how I think about it, let me know if it is right or wrong. I already know $\arcosh(x/a)$ is defined for $\ge a$. That is what we already knew. we want to however define a function when $x\le -a$. That is, when $\frac{x}{a}\le -1$. So adding a negative in the argument allows us to do so: $\arcosh(-x/a)$. Now our answer is defined for $x\le -a$ AND it just a reflection across the y-axis. Notice that $\arcosh(-x/a)$ implies $x=-a\cosh(\theta)$.

Putting together everything I've said, I have defined $x=a\cosh(\theta)$ for $x\ge a$ and $x=-a\cosh(\theta)$ for $x\le -a$, which is different from what you have. Of course, implying that $dx=a\sinh(\theta) \,d\theta$ for $x\ge a$ and $dx=-a\sinh(\theta) \,d\theta$ for $x\le -a$

 

[I like Serena]

This thread has too many rooms!
I have moved your new problem to a new thread.

 

[I like Serena]

$$=-\ln\left| \sec(\theta)+\tan(\theta) \right|+C$$

From a right triangle, we obtain the value of $\sec(\theta)=x/a$ and $\tan(\theta)=\sqrt{x^2-a^2}/a$, which, in the 2nd quadrant, secant is negative.

$$=-\ln\left| -x+\sqrt{x^2-a^2}\right|+C$$

You've lost me.
You start with $\sec(\theta)$, then you substitute $x/a$ for it, to end up with $-x$.
I'm afraid that I can't follow what you're doing. (Doh)

Also, do all even trig function have two inverses? $y=\cos(\theta)=\cos(-\theta)$, therefore, $\arccos(y)$ and $-\arccos(y)$. (Whew)

They have infinitely many inverses. (Smirk)

Consider that:
$$y=\cos(\theta) \Rightarrow \theta = \pm \arccos y + 2\pi k$$
So for instance $2\pi - \arccos y$ is also an inverse.

 

[I like Serena]

I think I finally understand this, but when I read back, it's making less sense. Why are we saying $\theta <0$ and $\theta >0$? Should it not be $x>a$ and $x<-a$?

Isnt that the same thing? (Wondering)

If $\theta>0$, then we have:
$$x = a \text{ sgn }(\theta) \cosh(\theta) = a \cosh(\theta) > a \quad \Rightarrow \quad x > a$$
And if $\theta<0$, we have:
$$x = a \text{ sgn }(\theta) \cosh(\theta) = -a \cosh(\theta) < -a \quad \Rightarrow \quad x < -a$$

 

[Dethrone]

I think it would be helpful to start from the beginning, since I'm starting to forget what the problem is. We want to find $\int \frac{dx}{\sqrt{x^2-a^2}}$ by method of trig substitution. Let $x=a \sec(\theta)$. Constructing a right angled triangle, we have the adjacent side is $a$, hypotenuse is $x$, and opposite side is $\pm \sqrt{x^2-a^2}$. Although most textbooks leave out the $\pm$, it will be helpful when we observe different quadrants.

Applying our substitution, we obtain:
$$\int \frac{a \sec(\theta)\tan(\theta)}{a|\tan(\theta)|}d\theta$$

From our original problem, we already know the domain of the function splits into two regions, when $x>a$, and when $x<-a$.

When $x>a$, $a \sec(\theta))>a$ and solving the inequality gives us the region when $0<\theta <\pi/2$. It is no coincidence that $\tan (\theta)$ just happened to be positive in that region, or is it?

Our problem simplifies to:
$$=\int \sec(\theta) \,d\theta$$
$$=\ln\left| \sec(\theta) +\tan(\theta) \right|+C$$
Because tangent is positive, we take the positive root of our triangle, which is $\frac{\sqrt{x^2-a^2}}{a}$.
$$=\ln\left| \frac{x}{a} +\frac{\sqrt{x^2-a^2}}{a}\right|+C$$

When $x<a$, $a \sec(\theta)<a$ and solving the inequality gives us the region when $\pi/2<\theta <\pi$. Again it is no coincidence that $\tan (\theta)$ just happened to be negative in that region, or is it? It's cool how observing the domain of the original integrand tells us exactly when $\tan(\theta)$ would be positive or negative.

Our problem simplifies to:
$$=-\int \sec(\theta) \,d\theta$$
$$=-\ln\left| \sec(\theta) +\tan(\theta) \right|+C$$
Because tangent is negative, we take the negative root of our triangle, which is $\frac{-\sqrt{x^2-a^2}}{a}$.
$$=-\ln\left| \frac{x}{a} -\frac{\sqrt{x^2-a^2}}{a}\right|+C$$
By a series of manipulation that you demonstrated to me earlier on:
$$=\ln\left| \frac{x}{a} +\frac{\sqrt{x^2-a^2}}{a}\right|+C$$

They are identical. :D

Aside: The textbooks usually draw the "triangles" without the $\pm$, but I think that's because they usually neglect explaining the different quadrants. My thoughts are that we're just using the "triangle" as a medium to relate trig ratios; the sides don't have to be just "positive". What are your thoughts? It makes sense though, because the opposite is $\pm \sqrt{x^2-a^2}$. Tangent could very much be $\frac{\sqrt{x^2-a^2}}{a}$ or $-\frac{\sqrt{x^2-a^2}}{a}$ depending on domain.
I think you could also focus your attention on the integrand: $\sqrt{x^2-a^2}=\left| a\tan(\theta) \right|$. When $\tan(\theta)>0$, we get $\frac{\sqrt{x^2-a^2}}{a}$, and vice-versa.

 

[Dethrone]

Isnt that the same thing? (Wondering)

If $\theta>0$, then we have:
$$x = a \text{ sgn }(\theta) \cosh(\theta) = a \cosh(\theta) > a \quad \Rightarrow \quad x > a$$
And if $\theta<0$, we have:
$$x = a \text{ sgn }(\theta) \cosh(\theta) = -a \cosh(\theta) < -a \quad \Rightarrow \quad x < -a$$

I had a suspicion they were the same, but I just can't think of it in terms of $\theta$. That is because $\cosh(\theta)$ is defined for both positive and negative arguments, so it's not quite clear why we're separating them in cases.

When I look at the graph of the $\arcosh$, it is clear to me that there is only one branch when $x>a$. We need to define a branch when $x<-a$, but that's impossible because $\arcosh(\frac{x}{a})$ isn't defined for negative arguments, so I stick in a negative, and there we go: $\arcosh(-\frac{x}{a})$, or equivalently, $x = -\arcosh(\theta)$.

Would you have said that my answer was complete in the begin, when it was just defined for $x>a$? Because a function only have one inverse, once you have chosen one, be it the positive or negative, you don't have both. What are you thoughts? (Wondering)

 

[I like Serena]

They are identical. :D

Yay! (Dance)

Aside: The textbooks usually draw the "triangles" without the $\pm$, but I think that's because they usually neglect explaining the different quadrants. My thoughts are that we're just using the "triangle" as a medium to relate trig ratios; the sides don't have to be just "positive". What are your thoughts? It makes sense though, because the opposite is $\pm \sqrt{x^2-a^2}$. Tangent could very much be $\frac{\sqrt{x^2-a^2}}{a}$ or $-\frac{\sqrt{x^2-a^2}}{a}$ depending on domain.
I think you could also focus your attention on the integrand: $\sqrt{x^2-a^2}=\left| a\tan(\theta) \right|$. When $\tan(\theta)>0$, we get $\frac{\sqrt{x^2-a^2}}{a}$, and vice-versa.

Typically, you would verify be taking the derivative of your answer.

Then it suffices to consider the $+$ case and make an educated guess about the $-$ case.
Taking the derivative then confirms you have the right answer, and it also tells you what more to do (if anything) for the $-$ case. (Nerd)

 

[I like Serena]

I had a suspicion they were the same, but I just can't think of it in terms of $\theta$. That is because $\cosh(\theta)$ is defined for both positive and negative arguments, so it's not quite clear why we're separating them in cases.

When I look at the graph of the $\arcosh$, it is clear to me that there is only one branch when $x>a$. We need to define a branch when $x<-a$, but that's impossible because $\arcosh(\frac{x}{a})$ isn't defined for negative arguments, so I stick in a negative, and there we go: $\arcosh(-\frac{x}{a})$, or equivalently, $x = -\arcosh(\theta)$.

Would you have said that my answer was complete in the begin, when it was just defined for $x>a$? Because a function only have one inverse, once you have chosen one, be it the positive or negative, you don't have both. What are you thoughts? (Wondering)

My thoughts are to keep things simple.

Just do the positive case and assume everything behaves nicely.
Then, when you have an answer, verify by taking the derivative.
Finally, consider what you would need in your answer to make it work for the negative case. And verify by checking what that would mean if you take the derivative. (Smile)

 

[Dethrone]

It's just a bit confusing:

If we just want to use a hyperbolic substitution, then we need only consider the substitutions $a=\cosh(\theta)$ and $a=-\cosh(\theta)$ depending on domain.

But if you want to find a perfect antiderivative $\int \frac{dx}{\sqrt{x^2-a^2}}$, then while $\arcosh(a)$ (from $a=\cosh(\theta)$) will meet that criteria, $\arcosh(-a)$ (from $a=-\cosh(\theta)$) will not. But by differentiation the latter, we notice that we need to add an extra negative outside to make it work: $-\arcosh(-a)$, is that what you mean? (Wondering) But, then, how did you notice the fact that separating the domain by cases such as $\theta >0$ and $\theta <0$ will solve this problem? I don't see that from what I did above.

 

[I like Serena]

It's just a bit confusing:

If we just want to use a hyperbolic substitution, then we need only consider the substitutions $a=\cosh(\theta)$ and $a=-\cosh(\theta)$ depending on domain.

But if you want to find a perfect antiderivative $\int \frac{dx}{\sqrt{x^2-a^2}}$, then while $\arcosh(a)$ (from $a=\cosh(\theta)$) will meet that criteria, $\arcosh(-a)$ (from $a=-\cosh(\theta)$) will not. But by differentiation the latter, we notice that we need to add an extra negative outside to make it work: $-\arcosh(-a)$, is that what you mean? (Wondering)

Yep!

Alternatively you can write $\arcosh$ as $\ln$, after which you can add absolute value symbols, just like you would for $\int x\,dx = \ln|x| + C$. (Nerd)

But, then, how did you notice the fact that separating the domain by cases such as $\theta >0$ and $\theta <0$ will solve this problem? I don't see that from what I did above.

The function $\cosh$ is an even function that is not invertible.
Only by limiting its domain (to $\theta \ge 0$) can we make it invertible.
That leaves us with half of its domain that we can use for other purposes.
Using the $\text{sgn}$ function makes it work.

(Yawn)

 

[Dethrone]

I think I realized my confusion.
You see, when $\theta<0$, then $x=-a\cosh(\theta)$
Then our graph is the left branch of the bottom parabola. Take the inverse of it, by flipping it across $y=x$ and you get the bottom branch of the sideways parabola. But that's not the part we want! We want the top portion of the sideways parabola. This works out mathematically, but not working out graphically for me. But otherwise, I understand it now. :D

Spoiler

Yes, this thread is getting really annoying now, but I think I would have gotten this much faster if you were my private tutor (Cool). LaTeXing and waiting for replies can be really annoying. The whole time, I was just trying to piece all your comments together, and trying to come up with your conclusion from scratch. Sometimes when I learn more math, I come back and read my old threads and cringe. Maybe I'm getting ahead of myself for asking these questions because I haven't even been exposed to hyperbolic functions much at all. When I asked you this question initially, I just knew that they could be represented by exponentials, and I didn't even know what they looked like. (Giggle)

 

[I like Serena]

Spoiler

Yes, this thread is getting really annoying now, but I think I would have gotten this much faster if you were my private tutor (Cool). LaTeXing and waiting for replies can be really annoying. The whole time, I was just trying to piece all your comments together, and trying to come up with your conclusion from scratch. Sometimes when I learn more math, I come back and read my old threads and cringe. Maybe I'm getting ahead of myself for asking these questions because I haven't even been exposed to hyperbolic functions much at all. When I asked you this question initially, I just knew that they could be represented by exponentials, and I didn't even know what they looked like. (Giggle)

Good!

Yeah. I realize that this thread was pretty messy.
I believe that is due to the multiple lines of discussion that were executed simultaneously.
It was fun though! ;)