An integration problem using trigonometric substitution

  • Thread starter mech-eng
  • Start date
  • #1
763
11

Homework Statement:

I cannot deal with this integration. ##\int\sqrt{1+y^2}##

Relevant Equations:

##\int\sqrt{1+y^2}##
##tan\theta=3y##
This is the integral I try to take. ##\int\sqrt{1+9y^2}## and ##9y^2=tan^2\theta## so the integral becomes ##\int\sqrt{1+tan^2\theta}=\sqrt {sec^2\theta}##. Now I willl calculate dy.

## tan\theta=3y ## and ##y=\frac {tan\theta}3## and ##dy=\frac{1+tan^2\theta}3##

Here is where I can only reach. I can't go any further.

Thanks.
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,812
1,390
This is the integral I try to take. ##\int\sqrt{1+9y^2}## and ##9y^2=tan^2\theta## so the integral becomes ##\int\sqrt{1+tan^2\theta}=\sqrt {sec^2\theta}##. Now I willl calculate dy.

## tan\theta=3y ## and ##y=\frac {tan\theta}3## and ##dy=\frac{1+tan^2\theta}3##
You need a factor of ##d\theta## on the last part.

Here is where I can only reach. I can't go any further.
Why not? What's stopping you?
 
  • #3
tnich
Homework Helper
1,048
336
It seems to me that your integral should include a ##dy##. You can easily state ##dy## in terms of ##sec\theta##. You are almost there already.
Have you thought about the limits of integration?
 
  • Like
Likes SammyS
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,650
Maybe it is only me, but I would use a hyperbolic substitution rather than a trigonometric one ...
 
  • Like
  • Informative
Likes Pi-is-3, Klystron and ehild
  • #5
ehild
Homework Helper
15,540
1,907
I would try integration by parts
##\int(\sqrt{1+y^2}) (1)dy##
 
  • Like
Likes HappyS5
  • #6
763
11
##tan\theta=3y##; so ## \frac{1+tan^2d\theta}3=dy## which is also equal to ##\frac {sec^2\theta}3d\theta=dy##

Now I obtained ##d\theta## and I will write the integral transformation.

##\int{ \sqrt{1+tan^2\theta} \frac {1+tan^2\theta}3d\theta }##

##\int\sqrt{sec^2\theta}\frac{sec^2\theta}3 d\theta##

But here I am confused by this integral. I write everything. I cannot do next step.
 
Last edited:
  • #7
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
##tan\theta=3y##; so ## \frac{1+tan^2d\theta}3=dy## which is also equal to ##\frac {sec^2\theta}3d\theta=dy##

Now I obtained ##d\theta## and I will write the integral transformation.

##\int{ \sqrt{1+tan^2\theta} \frac {1+tan^2\theta}3d\theta }##

##\int\sqrt{sec^2\theta}\frac{sec^2\theta}3 d\theta##

But here I am confused by this integral. I write everything. I cannot do next step.
Well, if a trigonometric substitution gets you nowhere, why not take the advice of Ordruin in #4 and use a hyperbolic substitution instead?
 
  • #8
ehild
Homework Helper
15,540
1,907
##\int\sqrt{sec^2\theta}\frac{sec^2\theta}3 d\theta##

But here I am confused by this integral. I write everything. I cannot do next step.
Is not it
##\int\frac{|sec^3(\theta)|}{3} d\theta##
 
  • #9
763
11
Is not it
##\int\frac{|sec^3(\theta)|}{3} d\theta##
Yes, that is exactly right. But I can't see the next step. And integrating an absolute value is a little confusing to me.
 
  • #10
ehild
Homework Helper
15,540
1,907
Yes, that is exactly right. But I can't see the next step. And integrating an absolute value is a little confusing to me.
Just ignore the absolute value sign for the time being.
There is a method to integrate rational functions of sin(x), cos(x), tan(x), by expressing them in terms of tan(x/2), and then using the substitution u=tan(x/2), so x=2 arctan(u),
dx = 2/(1+u2)du.
You will get a rational function to integrate.
##\cos(x)=\frac{1-u^2}{1+u^2}##
##\sin(x)=\frac{2u}{1+u^2}##
(check!)
Now you have the integral
##\int {\frac{1}{\cos^3\theta}d\theta}##
which becomes ##\int{(\frac{1+u^2}{1-u^2})^3\frac{2}{1+u^2}du}##.
You are able to integrate it by decomposing into partial fractions.
This method is a bit complicated, so I suggest to use @Orodruin's hyperbolic substitution, or integrating by parts.
 
  • Like
Likes mech-eng
  • #11
tnich
Homework Helper
1,048
336
You can also integrate ##\sec^3\theta## by parts. This breaks it down into functions that you probably already know how to integrate.
 
  • Informative
  • Like
Likes HappyS5 and mech-eng
  • #12
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,650
@mech-eng You have reacted to the latest posts, does that mean that you have solved your issues? If so it is informative to tell us about it and if not we need to know where you are still stuck. I still suggest that the easiest route is a hyperbolic substitution.
 
  • #13
763
11
I will turn back. I still look at some concepts and complete my integration knowledge. Sorry for such delays. For exampl I don't know how to take integral of ##\sec^3\theta## by parts.
 
  • #14
34,295
5,933
I will turn back. I still look at some concepts and complete my integration knowledge. Sorry for such delays. For exampl I don't know how to take integral of ##\sec^3\theta## by parts.
Your textbook might have this integral, either in a table of integrals or as an example of using integration by parts.
 
  • #15
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
I will turn back. I still look at some concepts and complete my integration knowledge. Sorry for such delays. For exampl I don't know how to take integral of ##\sec^3\theta## by parts.
What I fail to understand is why, oh why, you refuse to look at the simplest solution, but insist, stubbornly, to stick to just about the hardest way so solve the problem! If your aim is to develop skills at integration, the absolutely best way is to learn to be flexible: if one way is too hard or is taking too long, try something else.

Several of us have suggested you look at hyberbolic substitution instead of a trigonometric one. I can even offer a hint: the identity ##\cosh^2 u - \sinh^2 u = 1## allows us to write ##1+y^2 = \cosh^2 u## when we take ##y = \sinh u.##
 
  • #16
763
11
There is a worked example in Calculus with Analytical Geometry by Howard Anton, 3rd edition. It introduces "reduction formulas" as this form.

##\int sec^nx~~dx=\frac {sec^{n-2}tanx}{n-1} + \frac{n-2}{n-1} \int sec^{n-2}xdx ##

And in a previous section, the author gives it as an exercise to derive the reduction formula.

https://books.google.com.tr/books?i...ng powers of secant and tangent anton&f=false
And I can't also figure out why my latex code does not work.

Using the formula: ## \frac13 \int sec^3 \theta=\frac13 \frac{sec\theta tan\theta}{n-1}+ \frac{n-2}{n-1} \int{sec \theta dx} ##
 
Last edited:
  • Like
Likes HappyS5
  • #17
tnich
Homework Helper
1,048
336
There is a worked example in Calculus with Analytical Geometry by Howard Anton, 3rd edition. It introduces "reduction formulas" as this form.

##\int sec^nx~~dx=\frac {sec^{n-2}tanx}{n-1} + \frac{n-2}{n-1} \int sec^{n-2}xdx ##

And in a previous section, the author gives it as an exercise to derive the reduction formula.

https://books.google.com.tr/books?i...ng powers of secant and tangent anton&f=false
And I can't also figure out why my latex code does not work.

Using the formula: ## \frac13 \int sec^3 \theta=\frac13 \frac{sec\theta tan\theta}{n-1}+ \frac{n-2}{n-1} \int{sec \theta dx} ##
That is getting you closer to an answer, though as others point out, it may be easier to use hyperbolic functions. If you continue this route, you are missing a factor of ##\frac 1 3## on the last term.
 
  • #18
763
11
I will arrive at a result but now the problem is that after ##tan\theta=3y## substitution, how can I express ##sec\theta## in terms of y?

Thank you.
 
  • #19
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,650
I will arrive at a result but now the problem is that after ##tan\theta=3y## substitution, how can I express ##sec\theta## in terms of y?

Thank you.
Trigonometric identities. How is sec related to tan?
 
  • #20
763
11
Trigonometric identities. How is sec related to tan?
Since sec=##\frac 1{cos}~~and~~tan=\frac {sin}{cos}## then ##tan={sinsec}## but I don't know how to use this for ##y=3tan \theta##
 
  • #21
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,650
No, you need a trigonometric identity involving just tan and sec. Hint: You have already used it in this thread.
 
  • #22
763
11
No, you need a trigonometric identity involving just tan and sec. Hint: You have already used it in this thread.
This is might be the identiy required: ##\sec^2\theta=1+tan^2\theta## but I can't use it to manipulate ##sec\theta tan\theta##. I need more guidance.
 
  • #23
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,650
but I can't use it to manipulate ##sec\theta tan\theta##
Why not? It is perfectly possible. You know what ##\sec\theta## is in terms of ##\tan\theta## and you know what ##\tan\theta## is in terms of ##y##, so where is the problem?
 
  • Informative
Likes mech-eng
  • #24
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,812
1,390
This is might be the identiy required: ##\sec^2\theta=1+tan^2\theta## but I can't use it to manipulate ##sec\theta tan\theta##. I need more guidance.
Sometimes if you can't see how to proceed, it's worth it just to try something and see where it leads. It seems like you didn't even give it a shot here.

Anyway, here's an easy way to figure out what ##\sec \theta## is in terms of ##y##. The substitution was ##\tan \theta = 3y = \frac{y}{1/3}##, which corresponds to the right triangle below.

Untitled.jpg

It's straightforward to write ##\sec\theta## as a ratio of two sides of the triangle.
 
  • Like
  • Informative
Likes mech-eng and Mark44
  • #25
5
3
I am a bit confused. Why are people substituting ##3y = tan(\theta)##

I would substitute ##y=tan(\theta)##, and the deivative is ##\frac{dy}{dx} = tan^2(\theta) + 1 = sec^2(\theta)##. The final setup ## dy = sec^2(\theta) d \theta##.

##\int \sqrt(1 + y^2) dy = \int \sqrt(1 + tan^2(\theta)) \cdot sec^2(\theta) d\theta = \int sec^3(\theta) d\theta##

Now, I would use a reduction formula or integration by parts to solve the problem futher. The reduction formula can be derived by the integration by parts method.
 

Related Threads on An integration problem using trigonometric substitution

Replies
8
Views
2K
Replies
16
Views
707
Replies
3
Views
3K
Replies
12
Views
6K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
3
Views
764
Replies
5
Views
3K
Replies
9
Views
2K
Top