MHB Integration with trig and hyperbolic substitutions

[Dethrone]

And you were sloppy with the integration constants. (Wasntme)

I knew that...I was just too lazy to mention it as well.

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'm a bit confused, mostly because I don't quite understand the sign function too well.

Let's start from where I understand. To begin, we use the substitution $x=a\cosh\left({\theta}\right)$, so $dx=a\sinh\left({\theta}\right)$. If $\theta<0$, then it would simply be undefined as $\cosh\left({\theta}\right)$ is always greater than or equal to $1$. How can you distinguish cases when $\theta$ is not defined in our substitution $x=a\cosh\left({\theta}\right)$?

Spoiler

I hope this post makes it on a new page. $\LaTeX$ is starting to really lag...(Tmi)

 

[I like Serena]

Let's start from where I understand. To begin, we use the substitution $x=a\cosh\left({\theta}\right)$, so $dx=a\sinh\left({\theta}\right)$. If $\theta<0$, then it would simply be undefined as $\cosh\left({\theta}\right)$ is always greater than or equal to $1$. How can you distinguish cases when $\theta$ is not defined in our substitution $x=a\cosh\left({\theta}\right)$?

It seems you have started mixing up $\cosh$ and $\arcosh$.
$\cosh\theta$ is perfectly well defined for $\theta<0$. (Wink)

It's just not helping us, since it won't let us "reach" negative $x$.
That's where the sign function comes in.

Spoiler

I hope this post makes it on a new page. $\LaTeX$ is starting to really lag...(Tmi)

Yep. It did! ;)

 

[Dethrone]

$x=a\cosh\left({\theta}\right)$

If $\theta<0$, then isn't $dx$ simply $a\sinh\left({\theta}\right)$ as well? For example, $x=a\sin\left({\theta}\right)$, then regardless whether $\theta<0 \text{ or } \theta>0$, we still have $dx=a\cos\left({\theta}\right)$.

 

[I like Serena]

$x=a\cosh\left({\theta}\right)$

If $\theta<0$, then isn't $dx$ simply $a\sinh\left({\theta}\right)$ as well? For example, $x=a\sin\left({\theta}\right)$, then regardless whether $\theta<0 \text{ or } \theta>0$, we still have $dx=a\cos\left({\theta}\right)$.

This is true (if you add $d\theta$ in the appropriate places).

 

[Dethrone]

Ok! I will start to be very, very careful with my notation. (Smirk)

If that is true, then why did you add the extra negative? As we said in the above post, that for $\theta<0$, then $dx=a\sinh(\theta)\,d\theta$.

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 like Serena]

Ok! I will start to be very, very careful with my notation. (Smirk)

I hope so.

If that is true, then why did you add the extra negative? As we said in the above post, that for $\theta<0$, then $dx=a\sinh(\theta)\,d\theta$.

That's because we want a substitution that covers all of the domain of $x$.
With $x=a \cosh \theta$ we're only covering half of the domain.
With $\theta <0$ we're duplicating the same part of the domain of $x$ as with $\theta>0$.
The sign function is a cheap trick to use negative $\theta$ for the part of the domain that is not covered yet.

 

[Dethrone]

Oh, so we're just "sticking" a negative in there, so it's not a result of differentiation. Ah! because $\arcosh\left({x}\right)$ is restricted to values of $\theta \ge 1$ or more generally, $\theta \ge a$, so with the extra negative, then we can have it defined for $\theta \le -a$, since $\arcosh\left({-\frac{\theta}{a}}\right)$ is defined for negative $\theta$.

I think I'm seeing it now...

Let's try an example now, the one I started in my original post!
$$\int \frac{1}{\sqrt{x^2-a^2}}\,dx$$
$$=\int \frac{a \text{ sgn($\theta$)} \cosh\left({\theta}\right)}{a\sinh\left({\theta}\right)} \,d \theta$$
$$=\int \text{ sgn($\theta$)} \coth\left({\theta}\right) \,d \theta$$

Correct so far?

 

[I like Serena]

Oh, so we're just "sticking" a negative in there, so it's not a result of differentiation. Ah! because $\arcosh\left({\theta}\right)$ is restricted to values of $\theta \ge 1$ or more generally, $\theta \ge a$, so with the extra negative, then we can have it defined for $\theta \le -a$, since $\arcosh\left({-\frac{\theta}{a}}\right)$ is defined for negative $\theta$.

Exactly!

I think I'm seeing it now...

Let's try an example now, the one I started in my original post!
$$\int \frac{1}{\sqrt{x^2-a^2}}\,dx$$
$$=\int \frac{a \text{ sgn($\theta$)} \cosh\left({\theta}\right)}{a\sinh\left({\theta}\right)} \,d \theta$$
$$=\int \text{ sgn($\theta$)} \coth\left({\theta}\right) \,d \theta$$

Correct so far?

Erm... I'm counting 2 mistakes so far. Let's start with just the substitution.
(I always advise to do a substitution as a separate step, since this is where most mistakes are made. (Nerd))
$$\int \frac{1}{\sqrt{x^2-a^2}}\,dx
=\int \frac{1}{\sqrt{(a\text{ sgn }\theta \cosh \theta)^2-a^2}}\,d(a\text{ sgn }\theta \cosh \theta)$$
What do you get in the next step? (Wondering)

 

[Dethrone]

So $x=a \text{ sgn($\theta$)}\cosh\left({\theta}\right)$, $dx=a \text{ sgn($\theta$)} \sinh\left({\theta}\right) \, d \theta$

$$\int \frac{a\text{ sgn($\theta$)}\sinh\left({\theta}\right)}{\left| a \right|\sqrt{\text{ sgn$^2$($\theta$)}\cosh^2\left({\theta}\right)-1}} \,d \theta$$

$a>0$, so it cancels out with the top.
Correct so far?

Thanks so much for the help! I think I might need a name change to "I like ILS" (Giggle)(Rofl)

EDIT: Forgot my "$\,d \theta$", but caught it just in time. (Whew) Brb dinner. (Pizza)

 

[I like Serena]

So $x=a \text{ sgn($\theta$)}\cosh\left({\theta}\right)$, $dx=a \text{ sgn($\theta$)} \sinh\left({\theta}\right) \, d \theta$

$$\int \frac{a\text{ sgn($\theta$)}\sinh\left({\theta}\right)}{\left| a \right|\sqrt{\text{ sgn$^2$($\theta$)}\cosh^2\left({\theta}\right)-1}}$$

$a>0$, so it cancels out with the top.
Correct so far?

Much better! ;)
(Erm... can you add a $d\theta$?)

Thanks so much for the help! I think I might need a name change to "I like ILS" (Giggle)(Rofl)

Mmm... that would make you ILI or ILY. (Mmm)

 

[Dethrone]

Back!

$$\int \frac{a\text{ sgn($\theta$)}\sinh\left({\theta}\right)}{\left| a \right|\sqrt{\text{ sgn$^2$($\theta$)}\cosh^2\left({\theta}\right)-1}} \,d \theta$$
$$=\int \frac{\text{ sgn($\theta$)}\sinh\left({\theta}\right)}{\sqrt{\text{ sgn$^2$($\theta$)}\cosh^2\left({\theta}\right)-1}} \,d \theta$$

The $\text{ sgn$^2$($\theta$)}$ is annoying me, so I'll separate it into cases.
1. If $\theta >0$, then:

$$=\int \frac{\sinh\left({\theta}\right)}{\left| \sinh\left({\theta}\right)\right|} \, d\theta$$

Notice that $x=a \text{ sgn($\theta$)}\cosh\left({\theta}\right)$, then because $\theta>0$, then the expression simplifies to $x=a\cosh\left({\theta}\right)$ and $\cosh^{-1}\left({\frac{x}{a}}\right)=\theta$. This means that the outputs of $\theta$ will always be greater that $a$, which is greater than $0$, so we can take out the absolute value sign.

Correct? I'll do case 2 after this one.

Mmm... that would make you ILI or ILY. (Mmm)

Btw, in "ILY" what's the "Y" stand for? (Wondering) Being called ILI or ILY already makes me sound smarter.

 

[I like Serena]

Back!

$$\int \frac{a\text{ sgn($\theta$)}\sinh\left({\theta}\right)}{\left| a \right|\sqrt{\text{ sgn$^2$($\theta$)}\cosh^2\left({\theta}\right)-1}} \,d \theta$$
$$=\int \frac{\text{ sgn($\theta$)}\sinh\left({\theta}\right)}{\sqrt{\text{ sgn$^2$($\theta$)}\cosh^2\left({\theta}\right)-1}} \,d \theta$$

The $\text{ sgn$^2$($\theta$)}$ is annoying me, so I'll separate it into cases.

We can simplify with $\text{sgn}^2(\theta) = 1$.
That's just saying that both $- \cdot - = +$, and $+ \cdot + = +$.

1. If $\theta >0$, then:

$$=\int \frac{\sinh\left({\theta}\right)}{\left| \sinh\left({\theta}\right)\right|} \, d\theta$$

Notice that $x=a \text{ sgn($\theta$)}\cosh\left({\theta}\right)$, then because $\theta>0$, then the expression simplifies to $x=a\cosh\left({\theta}\right)$ and $\cosh^{-1}\left({\frac{x}{a}}\right)=\theta$. This means that the outputs of $\theta$ will always be greater or equal to zero, so we can take out the absolute value sign.

Correct? I'll do case 2 after this one.

Correct. :)

Btw, in "ILY" what's the "Y" stand for? (Wondering) Being called ILI or ILY already makes me sound smarter.

I Like You.

... and I'm going to sleep now. (Sleepy)

 

[Dethrone]

Well, answer this after you sleep. (Sleepy)

Yes, knowing $\text{sgn}^2(\theta) = 1$ will save me a lot of time, so I'll backtrack one step :D
$$\int \frac{\text{sgn}^2(\theta) \sinh\left({\theta}\right)}{\left| \sinh\left({\theta}\right) \right|} \,d \theta$$

1. Case 1: If $\theta>0$, then:
$$=\int 1 \,d \theta=\theta +C=\cosh^{-1}\left({\frac{x}{a}}\right)+C=\ln\left(\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right)+C$$
2. Case 2: If $\theta<0$, then:

Since $x=a\text{ sgn}(\theta)\cosh\left({\theta}\right)$ simplifies to $x=-a\cosh\left({\theta}\right)$ when $\theta <0$, then $\cosh^{-1}\left({-\frac{x}{a}}\right)=\theta$

$$=\int -1 \,d \theta=-\theta +C=-\cosh^{-1}\left({-\frac{x}{a}}\right)+C=\ln\left(-\frac x a - \sqrt{\left(\frac x a\right)^2-1}\right)+C$$

The moment I've been waiting for: amalgamation...

$$\int \frac{1}{\sqrt{x^2-a^2}} =\ln\left|\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right| + C$$
.
What do you think? (Wondering)

By the way...I just graphed $\arcosh\left({x}\right)$ on W|A, and although it's only defined when $x>1$, which it says so too on the page, how come it plots the section where $x<-1$? (Wondering)

y'='arcosh'('x')'–Wolfram|Alpha Clip 'n Share

This is the most mentally taxing and complicated question I've asked, and it's also the coolest. (Cool)

Also: integration - Evaluating the following integral: $ \int \frac{x^2}{\sqrt{x^2 - 1}} \text{ d}x$ - Mathematics Stack Exchange

Someone else also suggested the sign function! (There are some really bad advices on that thread...one does not simply ignore the absolute value...) They apply it a bit differently, and I do have a few questions regarding that, so maybe we can discuss that after this.

 

[I like Serena]

Well, answer this after you sleep. (Sleepy)

Yes, knowing $\text{sgn}^2(\theta) = 1$ will save me a lot of time, so I'll backtrack one step :D
$$\int \frac{\text{sgn}^2(\theta) \sinh\left({\theta}\right)}{\left| \sinh\left({\theta}\right) \right|} \,d \theta$$

1. Case 1: If $\theta>0$, then:
$$=\int 1 \,d \theta=\theta +C=\cosh^{-1}\left({\frac{x}{a}}\right)+C=\ln\left(\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right)+C$$
2. Case 2: If $\theta<0$, then:

Since $x=a\text{ sgn}(\theta)\cosh^{-1}\left({\theta}\right)$ simplifies to $x=-a\cosh^{-1}\left({\theta}\right)$ when $\theta <0$, then $\cosh^{-1}\left({-\frac{x}{a}}\right)=\theta$

$$=\int -1 \,d \theta=-\theta +C=-\cosh^{-1}\left({-\frac{x}{a}}\right)+C=\ln\left(-\frac x a - \sqrt{\left(\frac x a\right)^2-1}\right)+C$$

The moment I've been waiting for: amalgamation...

$$\int \frac{1}{\sqrt{x^2-a^2}} =\ln\left|\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right| + C$$
.
What do you think? (Wondering)

Looks good! ;)

By the way...I just graphed $\arcosh\left({x}\right)$ on W|A, and although it's only defined when $x>1$, which it says so too on the page, how come it plots the section where $x<-1$? (Wondering)

y'='arcosh'('x')'–Wolfram|Alpha Clip 'n Share

That's because W|A always works with complex numbers (usually denoted as $z$ instead of $x$).
If we have an $x<1$, it means that the result has to be complex, with an imaginary component between $\frac \pi 2$ and $\frac {3\pi}2$. (Nerd)

So suppose we want to calculate $\cosh^{-1} 0$, then:
$$e^{\cosh^{-1}(0)} = 0 + \sqrt{0^2-1} = i$$
Therefore
[box="green"]$$\cosh^{-1}(0) = \left(\frac \pi 2 + 2\pi k\right)i$$[/box]

This is the most mentally taxing and complicated question I've asked, and it's also the coolest. (Cool)

(Poolparty)(Smirk)

 

[Dethrone]

And like I've said, I want to follow-up with questions that I saw on stack exchange:
integration - Evaluating the following integral: $ \int \frac{x^2}{\sqrt{x^2 - 1}} \text{ d}x$ - Mathematics Stack Exchange

Now, what they did was they started $x = \cosh u$. Then they arrived at $\int \dfrac{\cosh^2 u \cdot \sinh u}{\left| \sinh u \right|} \text{ d}u$, which, in our question, is equivalent to $\int \dfrac{\sinh u}{\left| \sinh u \right|} \text{ d}u$ as we don't have the extra $x^2$ in our question.
Now they defined $|\sinh (u)| = \mathrm{sgn} (\sinh u) \sinh u$, so we now have $\int \mathrm{sgn}(\sinh (u)) du $, but wait a minute...
In the original substitution $\cosh^{-1} x=u$, meaning that it will only return values of $u>a>0$, implying that $\left| \sinh u \right|$ would always be positive. Why are they definition the sign function when that fact is already established? I don't see how that's working. (Wondering)

 

[I like Serena]

And like I've said, I want to follow-up with questions that I saw on stack exchange:
integration - Evaluating the following integral: $ \int \frac{x^2}{\sqrt{x^2 - 1}} \text{ d}x$ - Mathematics Stack Exchange

Now, what they did was they started $x = \cosh u$. Then they arrived at $\int \dfrac{\cosh^2 u \cdot \sinh u}{\left| \sinh u \right|} \text{ d}u$, which, in our question, is equivalent to $\int \dfrac{\sinh u}{\left| \sinh u \right|} \text{ d}u$ as we don't have the extra $x^2$ in our question.
Now they defined $|\sinh (u)| = \mathrm{sgn} (\sinh u) \sinh u$, so we now have $\int \mathrm{sgn}(\sinh (u)) du $, but wait a minute...
In the original substitution $\cosh^{-1} x=u$, meaning that it will only return values of $u>a>0$, implying that $\left| \sinh u \right|$ would always be positive. Why are they definition the sign function when that fact is already established? I don't see how that's working. (Wondering)

The return value $u$ is not necessarily at least zero.
We can also choose it be be negative.
Solving $\cosh u = x$ will give us both positive and negative values for $u$.
It's only $\arcosh x$ that is defined to be a function that only returns positive values.

Essentially that is what Thomas Andrews remarked upon - that we can pick $u$ as at least zero, so we can indeed do away with the sign function and the absolute function. (Nod)

 

[Dethrone]

The return value $u$ is not necessarily at least zero.
We can also choose it be be negative.
Solving $\cosh u = x$ will give us both positive and negative values for $u$.
It's only $\arcosh x$ that is defined to be a function that only returns positive values.
(Nod)

I'm slightly confused. What do you mean it's not "at least zero". And I thought the fact that $\arcosh x$ is defined to be a function that only returns positive values is all that matters.

Like in the past, with substitution $a=\sin x$, $x$ is not necessarily at least zero, it can be any number, but what returns the $x$ value, the $\arcsin u$ function, is all that matters when it comes to judging whether or not the absolute value should be kept. (The fact that $\arcsin u$ was restricted allowed us to remove the absolute value)

 

[I like Serena]

I'm slightly confused. What do you mean it's not "at least zero". And I thought the fact that $\arcosh x$ is defined to be a function that only returns positive values is all that matters.

It is not said that the substitution is $u = \arcosh x$. (Wait)

It could also be $u = -\arcosh x$.
Both of these substitutions fit $\cosh u = x$.
This is the choice that we can make.

 

[Dethrone]

Doesn't $u = -\arcosh x$ fit $\cosh(-u)=x$?
Oh, so in the cause of $x=\sin u$, we have both $u=\arcsin x$ and $u=-\arcsin x$, not that it matters, as they both outputs values between $\pi/2$ and $-\pi/2$. But I guess in this case it matters, because $u$ is now allowed to $<0$.

 

[I like Serena]

Doesn't $u = -\arcosh x$ fit $\cosh(-u)=x$?

Yes. And $\cosh(-u)=\cosh(u)$. (Smile)

Oh, so in the cause of $x=\sin u$, we have both $u=\arcsin x$ and $u=-\arcsin x$, not that it matters, as they both outputs values between $\pi/2$ and $-\pi/2$. But I guess in this case it matters, because $u$ is now allowed to $<0$.

That's different. Generally $\sin(u) \ne \sin(-u)$.
Instead we have $\sin(u) = -\sin(-u)$
So it does not apply here.

 

[Dethrone]

Oh! It's an even function! Nice... :D I get it...! But if we define the substitution as a signum or sign? function from the beginning, we can just amalgamate both cases: $u = \arcosh x$ and $u = -\arcosh x$ as $u= \text{ sgn(x)}\arcosh(x)$

Now I have a question about composition of hyperbolic functions:

So the question on stackexchange simplifies to this:
$$\frac 1 2u \, \mathrm{sgn}(\sinh u)+ \frac 1 4 \sinh (2u) \mathrm{sgn}(\sinh u)$$
Which is actually wrong because he forgot the "+C", but we will ignore that for now. (Giggle) I just want to focus on where $\sinh u>0$, since that's no longer confusing me.
$$=\frac{1}{2}\arcosh(x)\sinh(\arcosh(x))+...\text{etc}$$

How do I resolve the $\sinh(\arcosh(x))$? With circular trig, I would draw a triangle and apply the right angle rules, but I don't think this works for hyperbolic, or is there a right triangle for hyperbolic? (Wondering)

 

[I like Serena]

How do I resolve the $\sinh(\arcosh(x))$? With circular trig, I would draw a triangle and apply the right angle rules, but I don't think this works for hyperbolic, or is there a right triangle for hyperbolic? (Wondering)

There's probably an equivalent for a hyperbolic right triangle, but it makes my head hurt. (Doh)

So I'll stick (for now) with:
$$\sinh(\arcosh x) = \sinh\Big(\ln(x+\sqrt{x^2-1})\Big) = \frac 12 \Big(e^{\ln(x+\sqrt{x^2-1})} - e^{-\ln(x+\sqrt{x^2-1})}\Big) = \sqrt{x^2-1}$$
(Dull)

Darn! It does like some right triangle thingy! (Cool)

 

[I like Serena]

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:
$$y^2 = x^2 - 1$$
Therefore:

[hbox="blue"]$$y = \sinh(\arcosh x) = \sqrt{x^2 - 1}$$[/hbox]
(Mmm)

 

[Dethrone]

$$\frac 12 \Big(e^{\ln(x+\sqrt{x^2-1})} - e^{\ln(x+\sqrt{x^2-1})}\Big) = \frac 12 \left((x+\sqrt{x^2-1}) -(x+\sqrt{x^2-1}\right)=0?$$

There is some right triangle business going around (Smoking)...but I've seen a website where they rewrote all the hyperbolic functions in terms of the one they chose for their substitutions. One thing I've noticed is that $\cosh u=\sqrt{1+\sinh^2u}$ since $\cosh u$ is always positive...

So I guess you would recommend normal circular trig substitutions over hyperbolic because of all this unnecessary work? Which substitutions do you usually work with?

I just saw your post, I will read it now. :D

 

[I like Serena]

$$\frac 12 \Big(e^{\ln(x+\sqrt{x^2-1})} - e^{\ln(x+\sqrt{x^2-1})}\Big) = \frac 12 \left((x+\sqrt{x^2-1}) -(x+\sqrt{x^2-1}\right)=0?$$

There is some right triangle business going around (Smoking)...but I've seen a website where they rewrote all the hyperbolic functions in terms of the one they chose for their substitutions. One thing I've noticed is that $\cosh u=\sqrt{1+\sinh^2u}$ since $\cosh u$ is always positive...

I must be sliding off; I forgot a minus sign. (Blush)
Fixed now in my original post.

So I guess you would recommend normal circular trig substitutions over hyperbolic because of all this unnecessary work? Which substitutions do you usually work with?

I just use $\cosh x = \frac 12(e^x+e^{-x})$ and $\sinh x = \frac 12(e^x-e^{-x})$, which is almost identical as it is for $\cos$ and $\sin$.
I deduce anything else from those.

 

[I like Serena]

So I guess you would recommend normal circular trig substitutions over hyperbolic because of all this unnecessary work? Which substitutions do you usually work with?

Oh. I guess you mean which substitution to solve an integral? (Wondering)

Well, for the given integrals I'll just use whatever W|A or the table of integrals says I should use.
And if it doesn't quite fit, I think up a substition - any substitution that brings me closer.

For example, for $$\int \frac{dx}{\sqrt{x^2-a^2}}$$, no substitution is really necessary, since it's a standard integral.
Or with only the knowledge that $$\int \frac{dx}{\sqrt{u^2-1}} = \arcosh u + C$$, we need the substitution $$u=\frac x a$$. (Mmm)

 

[Dethrone]

But you're not allowed to consult W|A or an integral table during a test...I was referring to which method you use or had used when you couldn't consult them, such as when you were in university. (Wondering)

 

[I like Serena]

But you're not allowed to consult W|A or an integral table during a test...I was referring to which method you use or had used when you couldn't consult them, such as when you were in university. (Wondering)

At the time I knew the derivative of $\arcosh$ by heart. (Wasntme)

 

[Dethrone]

At the time I knew the derivative of $\arcosh$ by heart. (Wasntme)

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?

 

[I like Serena]

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)