Limits of integral after substitution

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  • #1
BruceW
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Hi everyone,
I am having a 'crisis of faith' in how the limits of an integral should change when you make a substitution for the variable involved. Especially when using a sinusoid substitution, since the sinusoidal functions are not 1-to-1 functions. Anyway, let's use an example integral:
[tex]\int_{r_0}^0 \sqrt{\frac{r}{r_0-r}} \ dr [/tex]
(where [itex]r_0[/itex] is a positive number). Now I want to use the substitution [itex] r = r_0 sin^2(\theta) [/itex] So the integral becomes:
[tex]\int \ 2r_0 sin^2(\theta) \ d \theta [/tex]
But what about the limits? for [itex]r=r_0[/itex], we need [itex]sin^2(\theta)=1[/itex], so we can choose theta as PI/2+n for any integer n. And we have a similar ambiguity for the other limit of [itex]r=0[/itex], we can choose any integer multiple of PI as our value of theta in the limit. So does this ambiguity allow us to reach different answers? Or can we rest easy knowing that we must always reach the same answer no matter what choice we make? Well, if we try PI/2 for the lower limit and zero for the upper limit we have:
[tex]\int_{\pi/2}^0 2r_0 sin^2(\theta) \ d \theta [/tex]
And using trigonometric identity, is equal to:
[tex]r_0 \int_{\pi/2}^0 1 \ d \theta - r_0 \int_{\pi/2}^0 cos(2\theta) \ d\theta [/tex]
And continuing the calculation, gives the answer to be:
[tex]- \frac{\pi r_0}{2} [/tex]
OK, so now trying it again, but instead use PI/2 for the lower limit and PI for the upper limit, we have:
[tex]\int_{\pi/2}^{\pi} 2r_0 sin^2(\theta) \ d \theta [/tex]
And what follows is a very similar calculation, which gives:
[tex]\frac{\pi r_0}{2} [/tex]
So it is the opposite sign of the other answer. So depending on the values of theta chosen for the limits, we get a different answer to the integral. But this is not what I would hope for, since we are not breaking any rules, the upper limit still corresponds to the same value of r in both cases. In practice, I know the negative answer is correct because I know the answer to the original integral. But if I didn't know the answer to the original integral, then I would be totally lost as to which answer is the correct one.

So, have I totally missed something... Or is there some rule that I have not followed? Many thanks in advance!
 
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Answers and Replies

  • #2
Curious3141
Homework Helper
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Hi everyone,
I am having a 'crisis of faith' in how the limits of an integral should change when you make a substitution for the variable involved. Especially when using a sinusoid substitution, since the sinusoidal functions are not 1-to-1 functions. Anyway, let's use an example integral:
[tex]\int_{r_0}^0 \sqrt{\frac{r}{r_0-r}} \ dr [/tex]
(where [itex]r_0[/itex] is a positive number). Now I want to use the substitution [itex] r = r_0 sin^2(\theta) [/itex] So the integral becomes:
[tex]\int \ 2r_0 sin^2(\theta) \ d \theta [/tex]
But what about the limits? for [itex]r=r_0[/itex], we need [itex]sin^2(\theta)=1[/itex], so we can choose theta as PI/2+n for any integer n. And we have a similar ambiguity for the other limit of [itex]r=0[/itex], we can choose any integer multiple of PI as our value of theta in the limit. So does this ambiguity allow us to reach different answers? Or can we rest easy knowing that we must always reach the same answer no matter what choice we make? Well, if we try PI/2 for the lower limit and zero for the upper limit we have:
[tex]\int_{\pi/2}^0 2r_0 sin^2(\theta) \ d \theta [/tex]
And using trigonometric identity, is equal to:
[tex]r_0 \int_{\pi/2}^0 1 \ d \theta - r_0 \int_{\pi/2}^0 cos(2\theta) \ d\theta [/tex]
And continuing the calculation, gives the answer to be:
[tex]- \frac{\pi r_0}{2} [/tex]
OK, so now trying it again, but instead use PI/2 for the lower limit and PI for the upper limit, we have:
[tex]\int_{\pi/2}^{\pi} 2r_0 sin^2(\theta) \ d \theta [/tex]
And what follows is a very similar calculation, which gives:
[tex]\frac{\pi r_0}{2} [/tex]
So it is the opposite sign of the other answer. So depending on the values of theta chosen for the limits, we get a different answer to the integral. But this is not what I would hope for, since we are not breaking any rules, the upper limit still corresponds to the same value of r in both cases. In practice, I know the negative answer is correct because I know the answer to the original integral. But if I didn't know the answer to the original integral, then I would be totally lost as to which answer is the correct one.

So, have I totally missed something... Or is there some rule that I have not followed? Many thanks in advance!
Hi Bruce! :smile:

There's a bit of a subtlety here. When you do the sub, this is what actually happens:

[tex]\int_{r_0}^0 \sqrt{\frac{r}{r_0-r}} \ dr = \int \sqrt{\frac{r_0\sin^2{\theta}}{r_0\cos^2{\theta}}}(2r_0)\sin\theta\cos{\theta} \ d{\theta} = \int \frac{|\sin{\theta}|}{|\cos{\theta}|}(2r_0)\sin{\theta}\cos{\theta} \ d{\theta}[/tex]

(I removed the bounds in the subbed integral for simplicity). Note that absolute value notation there, it's critical. Basically, the only way you're able to cancel out the cosine and simplify the whole expression is to take the case where sine and cosine have the same positive sign, so that implies a theta in the first quadrant.

Is that a satisfactory explanation?
 
  • #3
BruceW
Homework Helper
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Ah! yes. very good explanation, thank you! ah, that's awesome. my faith has been restored. I must try to keep on guard for those kinds of subtleties in the future. And so, I can rest assured that in similar problems, I can always choose any of the allowed values for theta, and the integral still comes out the same? (which is as I had hoped). And the thing that tripped me up here was actually something slightly different, that taking the square root of cos^2 gives the absolute value of cos, not cos itself. (a rookie mistake). Good, I think I understand it now.
 

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