Integration by substitution question

[doktorwho]

Homework Statement

Question:
To solve the integral ##\int \frac{1}{\sqrt{x^2-4}} \,dx## on an interval ##I=(2,+\infty)##, can we use the substitution ##x=\operatorname {arcsint}##?
Explain

Homework Equations

3. The Attempt at a Solution [/B]
This is my reasoning, the function ##\operatorname {arcsint}## can have values of ##t \in (-1,1)## because t doesn't fall anywhere else from that range and then follows that ##x \in (-\pi/2, +\pi/2)## which are the end values of the interval of ##t##. As the max point of intevral of ##x## is ##+\pi/2## which is roughly 1.57, this substitution can't be made as it doesn't even enter the original interval. Is this correct?
I have a couple of follow up question that I'm concerned about. I would be really glad if you can help me with them.
Say that the interval of ##x## that we got has some of its values from the original interval but not all, could we make the substitution then or only when they are all from there? Any special cases?
Thanks very much :)

 

[Mark44]

Homework Statement

Question:
To solve the integral ##\int \frac{1}{\sqrt{x^2-4}} \,dx## on an interval ##I=(2,+\infty)##, can we use the substitution ##x=\operatorname {arcsint}##?

I wouldn't use this substitution. Instead, I would use ##2\sec(\theta) = x##.

In a trig substitution I've always found it helpful to draw a right triangle, and label the two legs and hypotenuse in accordance with the expressions in your integral.
In my drawing, the hypotenuse is x, and the side adjacent to the angle is 2. The side opposite the angle is ##\sqrt{x^2 - 4}##. This gives ##\cos(\theta) = \frac 2 x##, or equivalently, ##2\sec(\theta) = x##. Use this relationship to convert expressions in x to ones in ##\theta## and to get ##d\theta## in terms of dx.

Explain

Homework Equations

3. The Attempt at a Solution [/B]
This is my reasoning, the function ##\operatorname {arcsint}## can have values of ##t \in (-1,1)## because t doesn't fall anywhere else from that range and then follows that ##x \in (-\pi/2, +\pi/2)## which are the end values of the interval of ##t##. As the max point of intevral of ##x## is ##+\pi/2## which is roughly 1.57, this substitution can't be made as it doesn't even enter the original interval. Is this correct?
I have a couple of follow up question that I'm concerned about. I would be really glad if you can help me with them.
Say that the interval of ##x## that we got has some of its values from the original interval but not all, could we make the substitution then or only when they are all from there? Any special cases?
Thanks very much :)

 

[Ray Vickson]

Homework Statement

Question:
To solve the integral ##\int \frac{1}{\sqrt{x^2-4}} \,dx## on an interval ##I=(2,+\infty)##, can we use the substitution ##x=\operatorname {arcsint}##?
Explain

Homework Equations

3. The Attempt at a Solution [/B]
This is my reasoning, the function ##\operatorname {arcsint}## can have values of ##t \in (-1,1)## because t doesn't fall anywhere else from that range and then follows that ##x \in (-\pi/2, +\pi/2)## which are the end values of the interval of ##t##. As the max point of intevral of ##x## is ##+\pi/2## which is roughly 1.57, this substitution can't be made as it doesn't even enter the original interval. Is this correct?
I have a couple of follow up question that I'm concerned about. I would be really glad if you can help me with them.
Say that the interval of ##x## that we got has some of its values from the original interval but not all, could we make the substitution then or only when they are all from there? Any special cases?
Thanks very much :)

Do you know if the integral is convergent? If it is, then using a change of variable may be sensible; if t is not (that is, if it diverges) no change of variable can be of any use at all (except maybe to clarify whether or not the integral converges).

 

[doktorwho]

I wouldn't use this substitution. Instead, I would use ##2\sec(\theta) = x##.

In a trig substitution I've always found it helpful to draw a right triangle, and label the two legs and hypotenuse in accordance with the expressions in your integral.
In my drawing, the hypotenuse is x, and the side adjacent to the angle is 2. The side opposite the angle is ##\sqrt{x^2 - 4}##. This gives ##\cos(\theta) = \frac 2 x##, or equivalently, ##2\sec(\theta) = x##. Use this relationship to convert expressions in x to ones in ##\theta## and to get ##d\theta## in terms of dx.

Do you know if the integral is convergent? If it is, then using a change of variable may be sensible; if t is not (that is, if it diverges) no change of variable can be of any use at all (except maybe to clarify whether or not the integral converges).

The question here is not to determine the solution to the integral but rather to show understanding of integrals and interval on which the solution is found. So its the question about whether this substitution would be possible.

 

[Ray Vickson]

The question here is not to determine the solution to the integral but rather to show understanding of integrals and interval on which the solution is found. So its the question about whether this substitution would be possible.

Yes, I know that was the question, but I could not make sense of your answer. Simple question: did you answer yes, or no?

 

[doktorwho]

Yes, I know that was the question, but I could not make sense of your answer. Simple question: did you answer yes, or no?

I answered no because when we make that substitution our x is in the interval ##(-\pi/2,\pi/2)## and that doesn't even belong to the original interval?

 

[Mark44]

Aside from any concern about intervals and definite integrals, why would anyone use ##x = \arcsin(t)## for this problem? If ##x = \arcsin(t)## then you're working with a right triangle with one acute angle of measure t, an opposite side of length x, and a hypotenuse of length 1. How would you tie these relationships to an integral that involves ##\sqrt{x^2 - 4}##?

 

[doktorwho]

Aside from any concern about intervals and definite integrals, why would anyone use ##x = \arcsin(t)## for this problem? If ##x = \arcsin(t)## then you're working with a right triangle with one acute angle of measure t, an opposite side of length x, and a hypotenuse of length 1. How would you tie these relationships to an integral that involves ##\sqrt{x^2 - 4}##?

The teacher asks these sort of things regarding substitutions to make us think about intervals on which they are defined. It doesn't have to be a reasonable substitution as we are not solving it anyway. Just to make us think about intervals.

 

[Mark44]

The teacher asks these sort of things regarding substitutions to make us think about intervals on which they are defined. It doesn't have to be a reasonable substitution as we are not solving it anyway. Just to make us think about intervals.

I agree that it's a good idea to think about the interval on which a substitution is defined, but it should go without saying that the substitution actually makes sense in relation to the integral. In this case, as I have already pointed out, the substitution makes no sense, so what's the use of investigating an interval?

If the integral had been ##\int \frac{dx}{\sqrt{1 - x^2}}##, then the given substitution would have been a reasonable substitution, and it would make sense to talk about the interval on which the substitution is defined.
It seems possible to me that your teacher didn't actually ask the question that he intended to ask.