limits of integration

**MechEng** · June 30th 2010, 11:40 AM

For the problem:

$ \int_{1}^{2}{\frac{1}{x^2\sqrt{4x^2+9}} $

I am getting limits of:

$tan^{-1}(3)$

and

$tan^{-1}(\frac{3}{2})$

Respectively.

Are these limits correct? Is there something easier to work with?

**mr fantastic** · June 30th 2010, 03:05 PM

Originally Posted by MechEng

For the problem:

$ \int_{1}^{2}{\frac{1}{x^2\sqrt{4x^2+9}} $

I am getting limits of:

$tan^{-1}(3)$

and

$tan^{-1}(\frac{3}{2})$

Respectively.

Are these limits correct? Is there something easier to work with?

State the substitution you are using. Then, please show your working in getting the integral terminals.

The substitution $x = \frac{1}{u}$ probably leads to easier calculations.

**wonderboy1953** · June 30th 2010, 03:14 PM

Originally Posted by MechEng

For the problem:

$ \int_{1}^{2}{\frac{1}{x^2\sqrt{4x^2+9}} $

I am getting limits of:

$tan^{-1}(3)$

and

$tan^{-1}(\frac{3}{2})$

Respectively.

Are these limits correct? Is there something easier to work with?

If you reverse the steps (i.e. differentiate), what do you get?

**MechEng** · June 30th 2010, 04:53 PM

I am using the substitution:

$x = \frac{3}{2}tan\theta$
$dx = \frac{3}{2}sec^2\theta d\theta$

Which yields:

$\int_{tan^{-1}(3)}^{tan^{-1}(\frac{3}{2})}{\frac{1}{(\frac{3}{2}tan\theta)^2 3sec\theta}\frac{3}{2}sec^2\theta d\theta$

I believe I am evaluating correctly and arriving at $-\frac{\sqrt{4x^2+9}}{9x}$

It just seems that there should be a neater set of limits for this integral.

**mr fantastic** · June 30th 2010, 05:36 PM

Originally Posted by MechEng

I am using the substitution:

$x = \frac{3}{2}tan\theta$
$dx = \frac{3}{2}sec^2\theta d\theta$

Which yields:

$\int_{tan^{-1}(3)}^{tan^{-1}(\frac{3}{2})}{\frac{1}{(\frac{3}{2}tan\theta)^2 3sec\theta}\frac{3}{2}sec^2\theta d\theta$

I believe I am evaluating correctly and arriving at $-\frac{\sqrt{4x^2+9}}{9x}$

It just seems that there should be a neater set of limits for this integral.

Yes, that's the substitution I thought you were using. But you haven't shown how you got your integral terminals (which are wrong, by the way. Which is why I asked you to show how you got them. Because you would then probably realise they were wrong and see the mistakes you have made).

Note that when you are solving a definite integral by making a substitution, you do not have to back-substitute for u to get an integral in terms of x. It is much better for many reasons (such as less time, less chance of mistake etc.) to stay with the new variable.

So you should be substituting your (correct) integral terminals into the u-integral that you calculate rather than trying to solve it in terms of x and the original terminals. Otherwise, why even bother getting new integral terminals if you're never going to use them. But, if you're determined to do it this way, then the anti-derivative you state is correct, so why are you even worrying about the u-integral terminals. Just use the x-integral terminals in your answer.

And did you consider the alternative substitution I suggested at all?

**MechEng** · June 30th 2010, 06:00 PM

Ah, I did consider the substitution you recommended. However, the section I am working on is trigonometric substitution. As such, I feel it would be best to stick with the trig substitution here.

So, if I am understanding this correctly... if I get my anti-derivative in terms of x, then there is no need to calculate new terminals? In this case my answer would simply be $-\frac{5}{18}+\frac{\sqrt{13}}{3}$ ?

What is the purpose of finding new terminals in this case?

I got my new terminals as shown below:

$x=1$

$\theta=x$

$1=\frac{3}{2}tan\theta$

$tan\theta=\frac{2}{3}$

$\theta=tan^{-1}(\frac{2}{3})$

AND

$x=2$

$\theta=x$

$2=\frac{3}{2}tan\theta$

$tan\theta=\frac{4}{3}$

$\theta=tan^{-1}(\frac{4}{3})$

Ah... does this look a little more correct?

**mr fantastic** · June 30th 2010, 06:13 PM

Originally Posted by MechEng

Ah, I did consider the substitution you recommended. However, the section I am working on is trigonometric substitution. As such, I feel it would be best to stick with the trig substitution here.

So, if I am understanding this correctly... if I get my anti-derivative in terms of x, then there is no need to calculate new terminals? In this case my answer would simply be $-\frac{5}{18}-\frac{\sqrt{13}}{3}$ ? Mr F says: The second term is not correct. Take more care with the arithmetic.

What is the purpose of finding new terminals in this case? Mr F says: None.

I got my new terminals as shown below:

$x=1$

$\theta=x$

$1=\frac{3}{2}tan\theta$

$tan\theta=\frac{2}{3}$

$\theta=tan^{-1}(\frac{2}{3})$

AND

$x=2$

$\theta=x$

$2=\frac{3}{2}tan\theta$

$tan\theta=\frac{4}{3}$

$\theta=tan^{-1}(\frac{4}{3})$

Ah... does this look a little more correct? Mr F says: Yes.

..

**MechEng** · June 30th 2010, 06:27 PM

I just caught my miscalculation upon rereading my post.

My book shows the act of finding new terminals as an important step, yet displays all "final" anti-derivatives in terms of x. Is there a reason that they would do this?

Should finding my answer with respect to theta be easier than solving in terms of x?

When I look at the following:

$\frac{2}{9}(-csc\theta)$ from $tan^{-1}(\frac{4}{3})$ to $tan^{-1}(\frac{2}{3})$ the answer is not obvious to me. Is there another rule or identity that I should be using?

I am really struggling to understand all of this again, and I do appreciate your patience with my questions.

**mr fantastic** · June 30th 2010, 06:46 PM

Originally Posted by MechEng

I just caught my miscalculation upon rereading my post.

My book shows the act of finding new terminals as an important step, yet displays all "final" anti-derivatives in terms of x. Is there a reason that they would do this? Mr F says: Who knows? There is no good reason (and several bad ones), in my opinion.

Should finding my answer with respect to theta be easier than solving in terms of x?

When I look at the following:

$\frac{2}{9}(-csc\theta)$ from $tan^{-1}(\frac{4}{3})$ to $tan^{-1}(\frac{2}{3})$ the answer is not obvious to me. Is there another rule or identity that I should be using?

I am really struggling to understand all of this again, and I do appreciate your patience with my questions.

If $\tan \theta = \frac{4}{3}$ you should be able to use either the Pythagorean Identity or a right-triangle to get the value of $\sin \theta$ and hence $\text{cosec} \, \theta$ . Ditto if $\tan \theta = \frac{2}{3}$ .

Evaluating $- \frac{2}{9} \left[ \text{cosec} \, \theta \right]^{\tan^{-1} \frac{4}{3}}_{\tan^{-1} \frac{2}{3}}$ should then be simple.

(But as I've said, if you go back to x then there's no good reason for doing this).

**MechEng** · June 30th 2010, 07:54 PM

Wow, it was much simpler using the anti-derivative and terminals in terms of theta in this case.

Thank you for the clarification.

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