Integration with trig and hyperbolic substitutions

[Dethrone]

Suppose we want to find:

$$\int \frac{1}{\sqrt{x^2-a^2}}\,dx$$

Trig Substitution:

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

Hyperbolic Substitution:

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

I know this is super minor, but how are they equivalent when one has the absolute value and one doesn't? Some people would consider it wrong without the absolute values, so I'm curious.

 

[I like Serena]

Suppose we want to find:

$$\int \frac{1}{\sqrt{x^2-a^2}}\,dx$$

Trig Substitution:

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

Hyperbolic Substitution:

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

I know this is super minor, but how are they equivalent when one has the absolute value and one doesn't? Some people would consider it wrong without the absolute values, so I'm curious.

Hey Rido! ;)

Taken literally, they are both wrong.
An indefinite integral needs to have an integration constant.

Furthermore, the indefinite integral has different solutions depending on which part of the domain you are taking the integral.

A "nice" solution covers all of the domain.
However, whenever a domain is split in disconnected intervals, there will be independent solutions on each interval - that can also have different integration constants.

Take for instance:

[box="yellow"]$$\int \frac{dx}{x} = \begin{cases} \ln(x) + C_1 & \text{if } x> 0 \\ \ln(-x) + C_2& \text{if } x< 0\end{cases}$$[/box]

You can verify that this is correct, by taking the derivative of the right hand side.

Anyway, this integral is usually solved as:

[box="yellow"]$$\int \frac{dx}{x} = \ln|x| + C$$[/box]

But as you can see, this is not a complete solution.
We need to interpret it as $C$ being potentially different constants on the 2 disconnected parts of the domain. (Nerd)

With your hyperbolic substitution, you're not quite done yet. Part of the domain has been neglected, which can probably be covered by doing a slightly different hyperbolic substitution.

 

[Dethrone]

My bad, I tend to forget constants of integration when I'm $\LaTeX$ing.

Are you trying to say that the trig substitution is giving us a more complete solution because it covers the full domain and that the hyperbolic substitution just gives when $x>a$? If that is true, wouldn't the trig substitution be more preferred?

What is the other sightly different hyperbolic substitution that covers the domain $x<a$?

 

[I like Serena]

My bad, I tend to forget constants of integration when I'm $\LaTeX$ing.

$\LaTeX$'ing has such a bad influence on you?
It usually helps me to visualize what I'm doing, so I notice mistakes like that! ;)

Are you trying to say that the trig substitution is giving us a more complete solution because it covers the full domain and that the hyperbolic substitution just gives when $x>a$? If that is true, wouldn't the trig substitution be more preferred?

What is the other sightly different hyperbolic substitution that covers the domain $x<a$?

Which trig respectively hyperbolic substitution did you use exactly? (Wondering)

It seems you used $x=a\cosh \varphi$.
Note that this only works if $x\ge 0$, since $\cosh$ is always positive or zero (assuming $a>0$).
For $x<0$, you should use the substitution $x=-a\cosh\varphi$.

 

[Dethrone]

It's just that when I'm $\LaTeX$'ing, I tend to delegate all my focus and time on the formatting, so things like that slip by me.

So, similar to the logarithm function, I should put the absolute value on the final answer when I'm done? :
$$=\cosh^{-1}\left({\frac{x}{a}}\right)=\ln \left| x+\sqrt{x^2-a^2} \right|$$
So that it will be true no matter the domain? (It's weird, because people who hyperbolic sub leave out the absolute value (Wondering) ) Should I do this every time? i.e \(\displaystyle \tanh^{-1}\left({x}\right)=\left| \frac{1+x^2}{1-x^2} \right|\)

From your experience, are there any situations where you notice that hyperbolic substitutions work better than the normal trig ones, or do you have to check the integrand each time to see which simplifies it better?

 

[I like Serena]

So, similar to the logarithm function, I should put the absolute value on the final answer when I'm done? :
$$=\cosh^{-1}\left({\frac{x}{a}}\right)=\ln \left| x+\sqrt{x^2-a^2} \right|$$
So that it will be true no matter the domain? (It's weird, because people who hyperbolic sub leave out the absolute value (Wondering) ) Should I do this every time? i.e \(\displaystyle \tanh^{-1}\left({x}\right)=\left| \frac{1+x^2}{1-x^2} \right|\)

From your experience, are there any situations where you notice that hyperbolic substitutions work better than the normal trig ones, or do you have to check the integrand each time to see which simplifies it better?

Hold on. That is not the way to go. $\arcosh x$ is only defined (for real numbers) for $x\ge 1$.
And in that case we simply have:
$$\arcosh x = \ln(x+\sqrt{x^2-1})$$
(With or without absolute symbols.)

 

[Dethrone]

I noticed that too. Previously you said $\cosh\left({x}\right)$ is always positive or greater than zero, but it should be always greater than $1$...

Now I'm confused ever more

Our answer using the substitution $x=a\sin\left({\theta}\right)$ is defined for all $x$, but using the hyperbolic substitution $x=a\cosh\left({x}\right)$, our answer is defined only when $x>1$. Why is that?

 

[Dethrone]

Furthermore, our substitution is $x=a\cosh\left({\theta}\right)$, so shouldn't it be $\frac{x}{a}>1$ or $x>a$.

 

[I like Serena]

I noticed that too. Previously you said $\cosh\left({x}\right)$ is always positive or greater than zero, but it should be always greater than $1$...

Now I'm confused ever more

Our answer using the substitution $x=a\sin\left({\theta}\right)$ is defined for all $x$, but using the hyperbolic substitution $x=a\cosh\left({x}\right)$, our answer is defined only when $x>1$. Why is that?

Yeah. Sorry for that. At least what I said wasn't wrong. (Angel)

It doesn't really matter you know.
If you have $\sqrt{x^2-a^2}$, you need that $|x|\ge a$ (assuming $a>0$).
So typically you need to distinguish the cases that either $x\ge a$ or $x \le -a$.

 

[I like Serena]

Furthermore, our substitution is $x=a\cosh\left({\theta}\right)$, so shouldn't it be $\frac{x}{a}>1$ or $x>a$.

It should be $\frac{x}{a}\ge 1$ or $x\ge a$ with this choice for the substitution. (Nerd)

 

[Dethrone]

No need to be sorry. (Blush)
If we have $\sqrt{x^2-a^2}$, then $x^2-a^2 \ge 0 \implies x^2 \ge a^2 \implies x \ge a$, assuming that both $a$ and $x$ are positive. Why is $|x|\ge a$?

It should be $\frac{x}{a}\ge 1$ or $x\ge a$ with this choice for the substitution. (Nerd)

Can you explain this? (Cool)

EDIT: Somehow the quote changed on me..(Wondering) I thought I read "$\frac{x}{a}\ge 1$ or $x\ge ax \ge a$", unless my mind is playing tricks on me (Smoking)

 

[I like Serena]

No need to be sorry. (Blush)
If we have $\sqrt{x^2-a^2}$, then $x^2-a^2 \ge 0 \implies x^2 \ge a^2 \implies x \ge a$, assuming that both $a$ and $x$ are positive. Why is $|x|\ge a$?

Well, you're assuming that $x$ is positive.
What if it is negative? (Sweating)

Can you explain this? (Cool)

Equality is also allowed... (Wait)

 

[I like Serena]

EDIT: Somehow the quote changed on me..(Wondering) I thought I read "$\frac{x}{a}\ge 1$ or $x\ge ax \ge a$", unless my mind is playing tricks on me (Smoking)

It must be $\LaTeX$ that is playing tricks. (Giggle)

 

[Dethrone]

Right. If it was negative, then $|x|\ge a$ would be more appropriate. (Nod)
But in this question, we have $\arcosh x = \ln(x+\sqrt{x^2-a^2})$, not just $\sqrt{x^2-a^2}$, so the restriction would be just $x\ge a$

I'm not sure, maybe I'm super slow, or $\LaTeX$ is really affecting me...
As I said above, it is $\arcosh x = \ln(x+\sqrt{x^2-a^2})$ only when $x\ge a$, but the other answer, we have $=\ln \left| x+\sqrt{x^2-a^2} \right|$, which is not only when $x\ge a$, but true for all $x\in \Bbb{R}$. Still not seeing how they're the same. (Crying)

 

[I like Serena]

Right. If it was negative, then $|x|\ge a$ would be more appropriate. (Nod)
But in this question, we have $\arcosh x = \ln(x+\sqrt{x^2-a^2})$, not just $\sqrt{x^2-a^2}$, so the restriction would be just $x\ge a$

I'm not sure, maybe I'm super slow, or $\LaTeX$ is really affecting me...
As I said above, it is $\arcosh x = \ln(x+\sqrt{x^2-a^2})$ only when $x\ge a$, but the other answer, we have $=\ln \left| x+\sqrt{x^2-a^2} \right|$, which is not only when $x\ge a$, but true for all $x\in \Bbb{R}$. Still not seeing how they're the same. (Crying)

What is $\d {} x \arcosh(-x)$? (Wondering)

 

[Dethrone]

$$\d{}{x}\arcosh\left({-x}\right)=\d{}{x}\ln\left({-x+\sqrt{x^2-1}}\right)=\frac{-1}{\sqrt{x^2-1}}$$

 

[I like Serena]

$$\d{}{x}\arcosh\left({-x}\right)=\d{}{x}\ln\left({-x+\sqrt{x^2-1}}\right)=\frac{-1}{\sqrt{x^2-1}}$$

Next step.
What is:
$$\d{}{x}\left(-\arcosh({-x})\right)$$
Ring a bell? (Wondering)

 

[Dethrone]

$$\frac{1}{\sqrt{x^2-1}}$$
Yes something's ringing, but it's not because I've noticed something...it's because my brain is starting to malfunction. (Headbang)

 

[I like Serena]

$$\frac{1}{\sqrt{x^2-1}}$$. Yes something's ringing, but it's not because I've noticed something...it's because my brain is starting to malfunction. (Headbang)

Let's try it like this... (Wasntme)

\begin{aligned}
\int \frac{1}{\sqrt{x^2-a^2}}
&= \begin{cases}
\arcosh\left(\frac x a\right) &+& C_1& \text{if } x \ge a \\
-\arcosh\left(-\frac x a\right)&+&C_2 & \text{if } x \le -a \end{cases} \\
&=\begin{cases}
\ln\left(\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right) &+& C_1 & \text{if }x \ge a \\
-\ln\left(-\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right) &+& C_2 & \text{if } x \le -a \end{cases} \\
&=\begin{cases}
\ln\left(\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right) &+& C_1 & \text{if } x \ge a \\
\ln\left(-\frac x a - \sqrt{\left(\frac x a\right)^2-1}\right) &+& C_2 & \text{if } x \le -a \end{cases} \\
&=\ln\left|\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right| + C &
\end{aligned}

 

[Dethrone]

Oh, essentially both $\arcosh\left(\frac x a\right)$ and $-\arcosh\left(-\frac x a\right)$ have the same derivative no matter the domain.

However, a few things I need clarification. $-\arcosh\left(-\frac x a\right)$ has two negatives introduced. I understand the one in the brackets, but why is there another negative sign outside? I know that it's necessary so that the negative introduced by the chain rule cancels it out, but I don't understand how $x\le -a$ implies a negative $\arcosh$.

I understand everything else, now. :D

 

[I like Serena]

Oh, essentially both $\arcosh\left(\frac x a\right)$ and $-\arcosh\left(-\frac x a\right)$ have the same derivative no matter the domain.

However, a few things I need clarification. $-\arcosh\left(-\frac x a\right)$ has two negatives introduced. I understand the one in the brackets, but why is there another negative sign outside? I know that it's necessary so that the negative introduced by the chain rule cancels it out, but I don't understand how $x\le -a$ implies a negative $\arcosh$.

I understand everything else, now. :D

It's the other way around when integrating.
We try to find a function such that when differentiated we get the function we have to integrate. (Nerd)

For $x\ge a$ we already know it is $\arcosh\left(\frac x a \right)$.
So for $x \le -a$ we make an educated guess and try $\arcosh\left(- \frac x a \right)$, which makes sense because the argument to $\arcosh$ has to be positive.
Turns out we left with a minus sign that shouldn't be there.
So we add a minus sign and presto! (Whew)Still... how did you do the trig substitution exactly? (Wondering)

 

[Dethrone]

Hi ILS! :D

I get it now! So you want to find an antiderivative that works when $x\le a$, which required a bit of working backwards.

Also:
$$=-\ln\left(-\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right)$$
$$=\ln\left(-\frac x a - \sqrt{\left(\frac x a\right)^2-1}\right) $$

Is there a fast way to get from the first to second statement? I'm not seeing it (Smoking)

The trig substitution I used was $x=a\sec\left({\theta}\right)$, I think I might have said a different substitution earlier on by mistake. (Wasntme)
$$\int \frac{dx}{\sqrt{x^2-a^2}}$$
$$=\int \sec\left({\theta}\right) \,d \theta$$
Amalgamating the cases of different domains:
$$=\int \ln \left| {\sec\left({\theta}\right)+\tan\left({\theta}\right)} \right|\,d \theta$$
$$=\ln \left| x+\sqrt{x^2-a^2} \right|$$

For the hyperbolic substitution, why can't we also amalgamate the domains from the get-go as well and simply say the answer is $=\ln \left| x+\sqrt{x^2-a^2} \right|$? (Wondering)

 

[I like Serena]

Hi ILS! :D

Hey Rido! (Happy)

I get it now! So you want to find an antiderivative that works when $x\le a$, which required a bit of working backwards.

Also:
$$=-\ln\left(-\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right)$$
$$=\ln\left(-\frac x a - \sqrt{\left(\frac x a\right)^2-1}\right) $$

Is there a fast way to get from the first to second statement? I'm not seeing it (Smoking)

Well...
\begin{aligned}-\ln(y+\sqrt{y^2-1}) &= \ln\frac 1 {y+\sqrt{y^2-1}} \\
&= \ln \frac {y-\sqrt{y^2-1}} {(y+\sqrt{y^2-1})(y-\sqrt{y^2-1})} \\
&= \ln \frac {y-\sqrt{y^2-1}} {y^2-(y^2-1)} \\
&= \ln(y-\sqrt{y^2-1})\end{aligned}
I take it you were hoping for something faster? (Giggle)

The trig substitution I used was $x=a\sec\left({\theta}\right)$, I think I might have said a different substitution earlier on by mistake. (Wasntme)
$$\int \frac{dx}{\sqrt{x^2-a^2}}$$
$$=\int \sec\left({\theta}\right) \,d \theta$$

Let's see:
$$
\begin{aligned}\int \frac{dx}{\sqrt{x^2-a^2}}
= \int \frac{d(a\sec\theta)}{\sqrt{(a\sec\theta)^2-a^2}} \\
= \int \frac{a\sec\theta \tan\theta\,d\theta}{a\sqrt{\tan^2\theta}} \\
= \int \frac{\sec\theta \tan\theta\,d\theta}{|\tan\theta|} \\
\end{aligned}
$$

How did you get an amalgamated $\int \sec\left({\theta}\right) \,d \theta$? (Wondering)

Amalgamating the cases of different domains:
$$=\int \ln \left| {\sec\left({\theta}\right)+\tan\left({\theta}\right)} \right|\,d \theta$$
$$=\ln \left| x+\sqrt{x^2-a^2} \right|$$

How did you get here? (Wondering)

For the hyperbolic substitution, why can't we also amalgamate the domains from the get-go as well and simply say the answer is $=\ln \left| x+\sqrt{x^2-a^2} \right|$? (Wondering)

Well... we could substitute $x = a \text{ sgn}(\theta) \cosh(\theta)$, where $\text{sgn}$ is the sign function.
That would make it possible to amalgamate the solution immediately. (Angel)

Note that we need a substitution that covers the complete domain of $x$.
With $x=a\cosh\theta$ we're only covering the positive part of the domain, which is why we won't get to an amalgamated solution immediately.

 

[Dethrone]

Hey Rido!

Well...
\begin{aligned}-\ln(y+\sqrt{y^2-1}) &= \ln\frac 1 {y+\sqrt{y^2-1}} \\
&= \ln \frac {y-\sqrt{y^2-1}} {(y+\sqrt{y^2-1})(y-\sqrt{y^2-1})} \\
&= \ln \frac {y-\sqrt{y^2-1}} {y^2-(y^2-1)} \\
&= \ln(y-\sqrt{y^2-1})\end{aligned}
I take it you were hoping for something faster? (Giggle)

Yes (Crying)

Let's see:
$$
\begin{aligned}\int \frac{dx}{\sqrt{x^2-a^2}}
= \int \frac{d(a\sec\theta)}{\sqrt{(a\sec\theta)^2-a^2}} \\
= \int \frac{a\sec\theta \tan\theta\,d\theta}{a\sqrt{\tan^2\theta}} \\
= \int \frac{\sec\theta \tan\theta\,d\theta}{|\tan\theta|} \\
\end{aligned}
$$

How did you get an amalgamated $\int \sec\left({\theta}\right) \,d \theta$? (Wondering)

By definition of the inverse secant, $\theta$ is in the first or third quadrant and, therefore, $\tan\left({\theta}\right)>0$, so we can take out the absolute value "without loss of generality". (Cool)

Spoiler

First time I used that phrase in quotations..does it work :D?

Now the $\tan\left({\theta}\right)$ crosses out and I get $\int \sec\left({\theta}\right) \,d \theta$

Using a well-known cheap trick:

$$\int \sec\left({\theta}\right) \frac{\sec\left({\theta}\right)+\tan\left({\theta}\right)}{\sec\left({\theta}\right)+\tan\left({\theta}\right)}\,d \theta$$
$$\int \frac{\sec^2\left({x}\right)+\sec\left({x}\tan\left({x}\right)\right)}{\sec\left({\theta}\right)+\tan\left({\theta}\right)}\, d \theta$$

Let $\sec\left({\theta}\right)+\tan\left({\theta}\right)=w$

$$\int \frac{1}{w}\,dw $$

Isn't amalgamating the domain of this solution analogous to hyperbolic substitution we used? IF $w>0$, then $\ln\left({w}\right)$. If $w<0$, then $\ln\left({-w}\right)$. Amalgamating domains, $\int \frac{1}{w}=\ln\left| w \right|+C$. Wait...no...it isn't the same as the hyperbolic one. Couldn't I simply write on my test paper "doing the same for $x\le -a$, with the substitution $a=-\cosh\left({-\frac{x}{a}}\right)$, I get $=\ln\left|\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right| + C$. (Poolparty)

How did you get here?

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

Well... we could substitute $x = a \text{ sgn}(\theta) \cosh(\theta)$, where $\text{sgn}$ is the sign function.
That would make it possible to amalgamate the solution immediately.

Note that we need a substitution that covers the complete domain of $x$.
With $x=a\cosh\theta$ we're only covering the positive part of the domain, which is why we won't get to an amalgamated solution immediately.

Nicccceee! (Cool) One thing:

$$x = a \text{ sgn}(\theta) \cosh(\theta)=a\frac{\left| \cosh\left({\theta}\right) \right|}{\cosh\left({\theta}\right)}$$

No...that's not quite right...

Spoiler

How do can I multi-quote without having to copy and paste

[/QUOTE" each time? (Wondering) Btw, I've took out some of your smileys/emotions in the quotes because I've hit the cap of 10...(Rofl)

 

[Dethrone]

I thought it was weird that it seemed like I'm the only person on the internet to ask this question...so I searched up "$\text{ sgn(x)}\cosh\left({x}\right)$.

Landed on a textbook: Mathematics for Engineers and Scientists, 5th Edition - Alan Jeffrey - Google Books

Under $\arcosh\left({\frac{x}{a}}\right)$, I think they defined it for the full domain! I'm not quite sure though.

$$\arcosh\left({\frac{x}{a}}\right)=\ln\left({\frac{\left| x \right|}{\left| a \right|}}+\frac{\sqrt{x^2-a^2}}{\left| a \right|}\right)$$
For $\left| \frac{x}{a} \right|>1$
$$=> \frac{x}{a}>1 \text{ or} \frac{x}{a}<-1 $$
$$=> x>a \text{ or }x<-a$$

Thus covering both domains, or am I solving it wrong?

 

[I like Serena]

By definition of the inverse secant, $\theta$ is in the first or third quadrant and, therefore, $\tan\left({\theta}\right)>0$, so we can take out the absolute value "without loss of generality". (Cool)

Spoiler

First time I used that phrase in quotations..does it work :D?

Now the $\tan\left({\theta}\right)$ crosses out and I get $\int \sec\left({\theta}\right) \,d \theta$

That's not really WLOG...
It's quite valid for $\theta$ to be in the 2nd or 4th quadrant.
So for those $\theta$ the integral is simply wrong!

And yes, it's quite usual to "neglect" the other quadrants at first, and focus on the 1st quadrant. But for a general answer, you will still need to examine the other quadrants. (Nerd)

Using a well-known cheap trick:

$$\int \sec\left({\theta}\right) \frac{\sec\left({\theta}\right)+\tan\left({\theta}\right)}{\sec\left({\theta}\right)+\tan\left({\theta}\right)}\,d \theta$$
$$\int \frac{\sec^2\left({x}\right)+\sec\left({x}\tan\left({x}\right)\right)}{\sec\left({\theta}\right)+\tan\left({\theta}\right)}\, d \theta$$

Let $\sec\left({\theta}\right)+\tan\left({\theta}\right)=w$

$$\int \frac{1}{w}\,dw $$

Isn't amalgamating the domain of this solution analogous to hyperbolic substitution we used? IF $w>0$, then $\ln\left({w}\right)$. If $w<0$, then $\ln\left({-w}\right)$. Amalgamating domains, $\int \frac{1}{w}=\ln\left| w \right|+C$.

Wow!
I didn't know that trick, so I'm not so sure if it is well-known.
I don't consider it cheap, but dirty.
And yes, it certainly is a trick!

It's like saying "it's obvious that". (Bandit)

Ah well, at least I can't find a mistake in it.

Wait...no...it isn't the same as the hyperbolic one. Couldn't I simply write on my test paper "doing the same for $x\le -a$, with the substitution $a=-\cosh\left({-\frac{x}{a}}\right)$, I get $=\ln\left|\frac x a + \sqrt{\left(\frac x a\right)^2-1}\right| + C$. (Poolparty)

Of course you could.
And maybe you score a bonus point for being aware of the special cases.
Or maybe a fraction of a point is subtracted with the comment please write it out. ;)

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

That doesn't look quite right...

Nicccceee! One thing:

$$x = a \text{ sgn}(\theta) \cosh(\theta)=a\frac{\left| \cosh\left({\theta}\right) \right|}{\cosh\left({\theta}\right)}$$

No...that's not quite right...

Nope. It's not.

How do can I multi-quote without having to copy and paste [QUOTE="I like Serena, post: 6717678, member: 312166"][/QUOTE] each time? Btw, I've took out some of your smileys/emotions in the quotes because I've hit the cap of 10...

I usually move the closing quote to somewhere in the beginning. And then I click the quick button to quote after selecting a section of text.

There's a cap of 10? I've never noticed before! (Leaving out the smiley)

$$\arcosh\left({\frac{x}{a}}\right)=\ln\left({\frac{\left| x \right|}{\left| a \right|}}+\frac{\sqrt{x^2-a^2}}{\left| a \right|}\right)$$
For $\left| \frac{x}{a} \right|>1$
$$=> \frac{x}{a}>1 \text{ or} \frac{x}{a}<-1 $$
$$=> x>a \text{ or }x<-a$$

Thus covering both domains, or am I solving it wrong?

That doesn't look quite right, since $\arcosh$ is only defined for $x \ge 1$. (Wondering)

 

[Dethrone]

I guess most calculus students neglect the other domains...and that part where I said "by definition of..." I plagiarized from my textbook (Giggle). But this reasoning works sometimes, right? Say we have the substitution $x=a\sin\left({x}\right)$. Then $\sqrt{1-\sin^2\left({x}\right)}=\left| \cos\left({x}\right) \right|$. Now we say that since the arcsine function has a domain of $-\frac{\pi}{2} \text{ to } \frac{\pi}{2}$, then the value of cosine will always be positive, allowing us to remove the absolute value.

Okay, so arcsecx is defined in other quadrants, how do we proceed from here? I will do this on my future tests, maybe I get bonus marks or look smart :D I'll also use hyperbolic substitutions with the sign function. (Cool)

$$= \int \frac{\sec\theta \tan\theta\,d\theta}{|\tan\theta|}$$

Wait...so we split the domain into two parts, when tangent is greater than zero, and when tangent in less than zero. The answer I had before is only one part of the answer!

________________________

Let me fix the answer:

$$\int \frac{1}{w}\,dw= \ln\left| w \right|= \ln \left| {\sec\left({\theta}\right)+\tan\left({\theta}\right)} \right|\,d \theta=\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$$

Looks good? I guess this is only correct for the first and 3rd quadrant...I will have to look at the other two quadrants...___________________________
How can I rewrite this in cases?
$$x = a \text{ sgn}(\theta) \cosh(\theta)$$_______________________

The textbook said that this

$$\arcosh\left({\frac{x}{a}}\right)=\ln\left({\frac{\left| x \right|}{\left| a \right|}}+\frac{\sqrt{x^2-a^2}}{\left| a \right|}\right)$$
is correct for $\left| \frac{x}{a} \right|>1$.

What was wrong with my deduction?
$=> \frac{x}{a}>1 \text{ or} \frac{x}{a}<-1$?

If $\left| x \right|>a$, then $x>a$ or $x<-a$

Spoiler

Asking questions via internet is tough! I think this could have been resolved much easier if you were my tutor in real life (Cool)

 

[I like Serena]

I guess most calculus students neglect the other domains...and that part where I said "by definition of..." I plagiarized from my textbook (Giggle). But this reasoning works sometimes, right? Say we have the substitution $x=a\sin\left({x}\right)$. Then $\sqrt{1-\sin^2\left({x}\right)}=\left| \cos\left({x}\right) \right|$. Now we say that since the arcsine function has a domain of $-\frac{\pi}{2} \text{ to } \frac{\pi}{2}$, then the value of cosine will always be positive, allowing us to remove the absolute value.

Hmm. arcsine does not have that domain...
It does have the range ($-\pi, \pi$].
In particular that means that you cannot leave out the absolute value.
Sorry.

Edit: That's a mistake.

Okay, so arcsecx is defined in other quadrants, how do we proceed from here? I will do this on my future tests, maybe I get bonus marks or look smart :D I'll also use hyperbolic substitutions with the sign function. (Cool)

$$= \int \frac{\sec\theta \tan\theta\,d\theta}{|\tan\theta|}$$

Wait...so we split the domain into two parts, when tangent is greater than zero, and when tangent in less than zero. The answer I had before is only one part of the answer!

Yep! (Happy)

Let me fix the answer:

$$\int \frac{1}{w}\,dw= \ln\left| w \right|= \ln \left| {\sec\left({\theta}\right)+\tan\left({\theta}\right)} \right|\,d \theta=\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$$

Looks good? I guess this is only correct for the first and 3rd quadrant...I will have to look at the other two quadrants...

I would write:
$$\int \frac{1}{w}\,dw= \ln|w| + C= \ln | {\sec(\theta)+\tan(\theta)}| + C
=\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$$
What are the differences? (Wondering)

How can I rewrite this in cases?
$$x = a \text{ sgn}(\theta) \cosh(\theta)$$

We can start without distinguishing cases:
$$x^2 = a^2 \cosh^2(\theta)$$
$$dx = a \text{ sgn}(\theta)\sinh(\theta)\,d\theta$$
How far can you get with that? (Wondering)

The textbook said that this

$$\arcosh\left({\frac{x}{a}}\right)=\ln\left({\frac{\left| x \right|}{\left| a \right|}}+\frac{\sqrt{x^2-a^2}}{\left| a \right|}\right)$$
is correct for $\left| \frac{x}{a} \right|>1$.

What was wrong with my deduction?
$=> \frac{x}{a}>1 \text{ or} \frac{x}{a}<-1$?

If $\left| x \right|>a$, then $x>a$ or $x<-a$

Say $x=-2a$, then $\left|\frac{x}{a}\right| > 1$, but $\arcosh(-2)$ is undefined.

Spoiler

Asking questions via internet is tough! I think this could have been resolved much easier if you were my tutor in real life (Cool)

(Cool) Quicker maybe, but posting helps to force one to think and review what is written.

 

[Dethrone]

Awesome! (Cool) The workbook I have been using is deceiving me :(, those are the explanations they give.

Wolfram Alpha says the range of the $\arcsin\left({x}\right)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}$, not $[-\pi, \pi]$, so does that revalidate my point? The thing is, every time an absolute value arises, I always question it, as you've seen with my previous threads. I did so with this too, but the workbook said it was okay to neglect.

_____________________________________________________________________
Quotes are too tedious to use, so I'll just use "_____________________".

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?

___________________________________________________________________
$$\int \frac{1}{w}\,dw= \ln|w| + C= \ln | {\sec(\theta)+\tan(\theta)}| + C
=\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$$

If this isn't right, nothing's right because I just copied and pasted your $\LaTeX$! But in all seriousness, I had an extra "$\,d \theta$"

__________________________________________________________________
$x^2 = a^2 \cosh^2(\theta)$

Why did you square both sides? (Wondering)
And how does taking the differential? of both sides give us this?
$dx = a \text{ sgn}(\theta)\sinh(\theta)\,d\theta$

__________________________________________________________________
Are you saying that the textbook is wrong to have said $\left| \frac{x}{a} \right|>1$, since $\arcosh(-2)$ is undefined?

 

[I like Serena]

Awesome! (Cool) The workbook I have been using is deceiving me :(, those are the explanations they give.

Wolfram Alpha says the range of the $\arcsin\left({x}\right)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}$, not $[-\pi, \pi]$, so does that revalidate my point? The thing is, every time an absolute value arises, I always question it, as you've seen with my previous threads. I did so with this too, but the workbook said it was okay to neglect.

Sorry. My mistake. The range of arcsine is [$-\frac\pi 2, \frac\pi 2$].
So cosine is always at least zero and you can leave out the absolute value.

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?

Yup. :D

$$\int \frac{1}{w}\,dw= \ln|w| + C= \ln | {\sec(\theta)+\tan(\theta)}| + C
=\ln \left| x+\sqrt{x^2-a^2} \right|+K, \text{ where K=-$\ln\left({a}\right)+C$}$$

If this isn't right, nothing's right because I just copied and pasted your $\LaTeX$! But in all seriousness, I had an extra "$\,d \theta$"

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

$x^2 = a^2 \cosh^2(\theta)$

Why did you square both sides? (Wondering)

Because that's how $x$ appears in the integral: as $x^2$.

And how does taking the differential? of both sides give us this?
$dx = a \text{ sgn}(\theta)\sinh(\theta)\,d\theta$

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)

Are you saying that the textbook is wrong to have said $\left| \frac{x}{a} \right|>1$, since $\arcosh(-2)$ is undefined?

As stated, it is indeed wrong.
Now if they would say for instance:
$$\arcosh\left|\frac{x}{a}\right|=...$$
Then it would be all right.

 

[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)