• Support PF! Buy your school textbooks, materials and every day products Here!

Arc length in calc 2

  • Thread starter LT72884
  • Start date
  • #1
143
7
Ello every one, i have interesting question. Any one who has james stewert 7th edition calc book im on secotion 8.1 studying for an exam.

number 13 of 8.1 says this

y= ln(secx)

find arc length from 0-pi/4

here is what i do first in my opinion.

y`= 1/sec(sectan)

y`= tanx

now i have √1+(y`)^2 = √1+tan^2

put it in an integral

∫√(1+tan^2)dx

according to books solution manual i do the following but i disagree

√1+tan^2 = √sec^2= secx

now they claim that the ∫secx = ln|secx + tanx| which i agree with. Now they say that is the answer HOWEVER, i say it is wrong because of one thing.

if the use a trig sub for back from chapter 7.3, they say you need to put a sub for dx as well which would be sec^2 so now the intergral is ∫sec^3x which is by parts...

do i not need to sub the dx because i did not start with 1+x^2 ??

thanks guys
 

Answers and Replies

  • #2
lurflurf
Homework Helper
2,432
132
I do not understand most of that.
We know that
$$\int \! \sec(x) \, \mathrm{dx}=\int \! \left( \frac{\tan(x/2)+1}{\tan(x/2)-1} \right)^{-1} \mathrm{d} \left( \frac{\tan(x/2)+1}{\tan(x/2)-1} \right) \, \mathrm{dx}=\log \left| \frac{\tan(x/2)+1}{\tan(x/2)-1} \right|=\log|\sec(x)+\tan(x)|$$
so the integral is correct. That is the trig sub as well.
What exactly do you disagree with?

I do not know what "put a sub for dx as well" means.
The chapter references to your terrible book is not helpful, I fortunately do not have one nearby.
 
  • #3
77
9
First of all, you have: y = ln(sec(x)).

Therefore, dy/dx = tan(x). Now you want to find 1+(dy/dx)², which is 1 + tan²(x) = sec²(x).

We know that L is equal to the integral √sec²(x) dx, with limits from 0 to π/4, which is pretty easy to evaluate.

What's the problem?
 
  • #4
143
7
you guys need to be a we bit more considerate when replying.

here is what i mean by a sub for dx.

according to my book when you do a trig sub dx must be changed into a trig function as well.

from the book section 7.3

∫√1+x^2

let a=1
let x=(a)tanx = tanx
dx=sec^2
x^2 = tan^2

so we have ∫√(1+tan^2)sec^2

now that becomes

∫sec^3

that is exactly from the book step by step. Now i must use by parts for the integration of sec^3. So my question is when i have the

∫√(1+tan^2)

for arc length, then why does it seem i get two different answers for the ∫√(1+tan^2)

is it simply because one of the problems, i am substituting x to be tan and the other one i am not substituting any x's??

that is what i am trying to understand.

thanks
 
  • #5
77
9
I don't understand why your book would do that. The integral of √sec²(x) dx is simply the integral of sec(x) dx, because √sec²(x) = |sec(x)| = sec(x), from 0 to π/4.
 
  • #6
lurflurf
Homework Helper
2,432
132
You need to be a we bit more considerate when posting unintelligible questions.
The two integrals you are comparing are
$$\int \! \sqrt{1+\tan^2(x)}\, \mathrm{d}\tan(x)=\int \! \sqrt{(1+\tan^2(x))^3}\, \mathrm{d}x=\int \! \sec^3(x) \, \mathrm{d}x \\
\text{and} \\
\int \! \frac{\mathrm{d}\tan(x)}{\sqrt{1+\tan^2(x)}} =\int \! \sqrt{1+\tan^2(x)}\, \mathrm{d}x=\int \! \sec(x) \, \mathrm{d}x$$

They are not the same.

If you prefer to write u=tan(x) we have

$$\int \! \sqrt{1+u^2} \, \mathrm{d}u=\int \! \sqrt{(1+u^2)^3}\, \mathrm{d}x=\int \! \sec^3(x) \, \mathrm{d}x \\
\text{and} \\
\int \! \frac{\mathrm{d}u}{\sqrt{1+u^2}} =\int \! \sqrt{1+u^2}\, \mathrm{d}x=\int \! \sec(x) \, \mathrm{d}x$$

In other words the correct way to compute by trig sub is

$$\int \! \sec(x) \, \mathrm{d}x=\int \! \frac{\mathrm{d}u}{\sqrt{1+u^2}}=\sinh^{-1}(u)=\sinh^{-1}(\tan(x))$$

also note that

$$1+\sqrt{2}=\sqrt{3+2 \sqrt{2}}$$

so the four answers are equivalent

$$\log(1+\sqrt{2}) \\ \log \left( \sqrt{3+2\sqrt{2}} \right) \\ \frac{1}{2}\log ( 3+2\sqrt{2} ) \\ \sinh^{-1}(1)$$
 
Last edited:
  • #7
33,505
5,191
here is what i mean by a sub for dx.

according to my book when you do a trig sub dx must be changed into a trig function as well.

from the book section 7.3

∫√1+x^2

let a=1
let x=(a)tanx = tanx
dx=sec^2
x^2 = tan^2
I'm certain your book doesn't say this; namely, "let x=(a)tanx = tanx"
Also, it wouldn't say this: "dx=sec^2"

It probably does something like this:
Let u = tan(x)
Then du = sec2(x)dx

When you make a substitution, you always want to use a different variable.
 
  • #8
143
7
here is where the confusion starts. haha. My book does specifically say the following.

if you have the expression

√(a^2 + x^2)

use x=atan and dx = sec^3

example from 7.3

∫√(1+x^2)dx

let x=atan
dx=sec^2
x^2=tan^2
tan=x/1

then book shows this:

∫√(1+tan^2)dx

book makes the sub for dx and then the inside like so:

∫√(1+tan^2)sec^2

then:

∫√(sec)sec^2

∫√(sec^3)

ok so that is section 7.3 haha. Mark, I'm not sure why they chose x rather than u. haha.

now in 8.1 one, it asks for the arc length for the following:

∫√(1+tan^2)dx.

ok sorry to repeat but where the confusion lies is now i have an integral that looks EXACTLY like the one from 7.1, however, it did not start out as ∫√(1+x^2).

It started out as ∫√(1+(y`)^2) and y`=tan

so it looks exactly the same as the 7.1 example but i dont think it is treated the same.. thats where the confusion lies. The book does use x and dx for substitutions as shown in 7.1. They also use u substitution as well.

it is knowing when to substitute dx with a trig sub and knowing when not to is part of my confusion.

thanks
 
Last edited:
  • #9
143
7
lurflurf, you are an a$$. If you do not like my "unintelligible questions" please feel free not to answer them. I came here for learning and understanding, not to be treated like a child.

To everyone else, thanks for at least trying to understand what i am talking about and the confusion i am receiving from my book. Next time, i will be asking other people rather than here. I do appreciate your time and consideration though. Thanks Mark for trying to help me out, I appreciate it.
 
  • #10
33,505
5,191
here is where the confusion starts. haha. My book does specifically say the following.

if you have the expression

√(a^2 + x^2)

use x=atan and dx = sec^3
No reputable book would write this.
It might have
Let x = a * tan(θ)
and dx = a * sec2(θ)dθ
example from 7.3

∫√(1+x^2)dx

let x=atan
dx=sec^2
x^2=tan^2
tan=x/1

then book shows this:

∫√(1+tan^2)dx

book makes the sub for dx and then the inside like so:

∫√(1+tan^2)sec^2
I don't believe your book has this. You are ignoring a lot of important information, such as the input argument to the tangent function, and the differentials. You can't write just tan or sec2; you have to include what you're taking the tangent of or the secant of.
then:

∫√(sec)sec^2

∫√(sec^3)
These should be:
∫√(sec(x))sec^2(x)dx
=∫√(sec^3(x))dx
ok so that is section 7.3 haha. Mark, I'm not sure why they chose x rather than u. haha.

now in 8.1 one, it asks for the arc length for the following:

∫√(1+tan^2)dx.
Closer, but still not quite write. It should be
∫√(1+tan^2(x))dx
ok sorry to repeat but where the confusion lies is now i have an integral that looks EXACTLY like the one from 7.1, however, it did not start out as ∫√(1+x^2).

It started out as ∫√(1+(y`)^2) and y`=tan

so it looks exactly the same as the 7.1 example but i dont think it is treated the same.. thats where the confusion lies. The book does use x and dx for substitutions as shown in 7.1. They also use u substitution as well.

it is knowing when to substitute dx with a trig sub and knowing when not to is part of my confusion.
Part of your confusion is a result of not noticing what is in your book and omitting some important details. For an integral like this: ##\int \sqrt{x^2 + 1}dx##, they are using the substitution tan(θ) = x/1, or x = tan(θ). Once you have this substitution, then take differentials to find dx, which in this case would be dx = sec2(θ)dθ.

Now replace x and dx in the original integral by θ and dθ according to the two equations above. This should give you an integral that's easier to evaluate.
 
  • #11
lurflurf
Homework Helper
2,432
132
Like I said before work out in both directions the substitutions

$$\int \! \sqrt{1+u^2} \, \mathrm{d}u=\int \! \sec^3(x) \, \mathrm{d}x \\
\text{and} \\
\int \! \frac{\mathrm{d}u}{\sqrt{1+u^2}} =\int \! \sec(x) \, \mathrm{d}x$$

You will see why they are not the same.

The book flawed as it is suggests the correct steps (keep in mind that the restricted domain 0<=x<=pi/4 is being used)

$$\int \! \sqrt{1+\tan^2(x)} \, \mathrm{d}x=\int \! \sqrt{\sec^2(x)} \, \mathrm{d}x=\int \! \sec(x) \, \mathrm{d}x=\log|\sec(x)+\tan(x)|+\mathrm{Constant}$$

There is not much to disagree with there. It disagrees with your wrong trig sub, but that is because the trig sub is wrong. Here are the trig subs both integrals.

$$\int \! \sqrt{1+\tan^2(x)} \, \mathrm{d}x=\int \! \sec(x) \, \mathrm{d}x \\
\text{write in easy to substitute form} \\
\int \! \sqrt{1+\tan^2(x)} \frac{\tan^{\prime}(x)}{1+\tan^2(x)} \, \mathrm{d}x=\int \! \sec(x) \frac{\sec^{\prime}(x) / \sec(x)}{\sqrt{\sec^2(x)-1}} \, \mathrm{d}x \\
\text{Substitute} \\ u=\tan(x) \\v=\sec(x) \\
\int \! \frac{\mathrm{d}u}{\sqrt{1+u^2}}=\int \! \frac{\mathrm{d}v}{\sqrt{v^2-1}} \\
\text{integrate} \\
\sinh^{-1}(u)+\mathrm{Constant}=\log \left| \sqrt{v^2-1}+v \right|+\mathrm{Constant} \\
\text{back substitute}\\ u=\tan(x) \\v=\sec(x) \\
\sinh^{-1}(\tan(x))+\mathrm{Constant}=\log \left| \sqrt{\sec^2(x)-1}+\sec(x) \right|+\mathrm{Constant} \\
\text{both of which are equal to}\\
\log|\sec(x)+\tan(x)|+\mathrm{Constant}
$$
 
  • #12
559
8
The book does not have what he has written... I just checked because I remember doing this exact problem last semester. The solution manual doesn't either they have it like this:
[tex]y=ln(sec x)\implies \frac{dy}{dx}=\frac{secxtanx}{secx}=tanx \implies 1+(\frac{dy}{dx})^{2}=1+tan^{2}x=sec^{2}x[/tex]
So: [tex]L=\int_{0}^{\frac{\pi}{4}} \sqrt{sec^{2}x}dx=\int_{0}^{\frac{\pi}{4}} |secx|dx=\int_{0}^{\frac{\pi}{4}} secxdx[/tex]
[tex]=ln(secx+tanx)|_{0}^{\frac{\pi}{4}}=ln(sqrt{2}+1)-ln(1+0)=ln(\sqrt{2}+1[/tex]
 
  • Like
Likes 1 person

Related Threads on Arc length in calc 2

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
6
Views
697
  • Last Post
Replies
3
Views
875
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
1
Views
1K
Replies
7
Views
8K
  • Last Post
Replies
5
Views
1K
Top