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This is the integral in order to made some approximation after the result. I need the result in term of $\arccos$. $$ \int_l^k\! \sqrt{\frac{R}{r} -1} \:\mathrm{d}r = \cos^{-1} (...). $$

Thomas Russell
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physnolimits
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  • I presume R is a constant? – User1234 Jun 19 '15 at 12:51
  • Yes R is a constant only r is a variable – physnolimits Jun 19 '15 at 13:50
  • This is not a math exercise I need it for a calculation in nuclear physics – physnolimits Jun 19 '15 at 14:06
  • @physnolimits If you just need the answer and the procedure does not matter, you can take Wolfram Alpha's solution and "convert" it with the procedure suggested by Rory Daulton. Though in one of the terms in Wolfram's solution the "conversion" is to put $\arccos(\cos(...))$. – Ian Jun 19 '15 at 14:10
  • The procedure matter, in fact wolfram goes straight to the point and doesn t work because i need a specific kind of result is not a definitive answer it s just a piece in the mosaic – physnolimits Jun 19 '15 at 14:12

2 Answers2

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Here's my suggestion. The minus sign suggests that the integrand would be the leg of some right triangle, where the other leg is $1$. Then the hypotenuse is $\sqrt{R/r}$. So we wind up with

$$\sqrt{R/r-1}=\tan(\theta) \\ r=R \cos(\theta)^2 \\ dr=-2R \cos(\theta) \sin(\theta) d \theta.$$

So your integral (limits aside for now) becomes

$$\int -2R \tan(\theta) \cos(\theta) \sin(\theta) d \theta = \int -2R \sin(\theta)^2 d \theta.$$

This should be easy to recognize; with power reduction you get

$$-R\theta + \frac{R\sin(2 \theta)}{2} + C = -R \theta + R \sin(\theta) \cos(\theta) + C.$$

Use the diagram I described in the first paragraph to simplify:

$$-R\arccos(\sqrt{r/R})+R\frac{\sqrt{R/r-1}}{\sqrt{R/r}}\frac{1}{\sqrt{R/r}} + C = -R \arccos(\sqrt{r/R})+r\sqrt{R/r-1} + C$$

Now you insert your limits. (I assume that $k,l \in [0,R]$.)

Also please check my work, because my answer seems quite different from Wolfram Alpha's answer.

Ian
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  • your work is good bu there is something that doesn't work later I'm checking all the rest, but your answer is the most similar to what I have to do... – physnolimits Jun 19 '15 at 14:35
  • @physnolimits There were several edits where I was performing simplifications, so be careful. – Ian Jun 19 '15 at 14:36
  • @physnolimits I think my answer is correct now, because I've now written it in a form where one of the terms is exactly the first term obtained when you do integration by parts (which is probably how Wolfram Alpha's answer begins, internally). – Ian Jun 19 '15 at 14:42
  • @physnolimits But if $R>1$ then this cannot be the arccosine of anything, unless there is some additional constraint on $k,l$ that you haven't specified. – Ian Jun 19 '15 at 14:47
  • k=R l is positive l<<R – physnolimits Jun 19 '15 at 14:50
  • @physnolimits Then as long as $R$ is a little bit bigger than $1$, your integral is not the arccos of anything (since it will be slightly less than $R\pi/2$). – Ian Jun 19 '15 at 14:51
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Here are some hints to help you on your way. To get more help you should show more of your own effort.

First, find any indefinite integral of $\sqrt{\frac Rr-1}$. You will probably get an answer that has an arctangent in it. Let's say that the answer is

$$I=v+c\tan^{-1}(u)$$

where $u$ and $v$ are functions of $r$ and $c$ is a constant.

Then use the standard right triangle diagram to convert an arctangent to an arccosine. You will get:

$$\theta=\tan^{-1}(u)\quad\implies\quad\theta=\cos^{-1}\left( \frac 1{\sqrt{1+u^2}} \right)$$

enter image description here

Therefore you can convert your integral answer to

$$I=v+c\cos^{-1}\left( \frac 1{\sqrt{1+u^2}} \right)$$

Is that clear?

The problem with that answer is that it uses the arccosine function but is not in its entirety an arccosine. If your answer to your definite integral is in the appropriate range, you could always use this trivial answer:

$$I=\cos^{-1}\left(\cos(I)\right)$$

If your answer to your definite integral is not in the appropriate range then there is no answer to your question.

Rory Daulton
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  • Sorry but isn't si clear to me I can't see the point – physnolimits Jun 19 '15 at 14:08
  • First get any answer for that integral: it will probably use the arctangent function. Have you gotten that far yet? What I wrote above will make it easy to convert that answer to one that uses the arccosine function. Isn't that what you want? – Rory Daulton Jun 19 '15 at 14:12
  • @Ian: I graphed your function and the one from WolframAlpha/Dr. Sonnhard Graubner and the graphs are different. As far as I can tell you did make an error. It is possible that I made a graphing error: did you do a graph check of your own? – Rory Daulton Jun 19 '15 at 14:24
  • @RoryDaulton Can you find it in my answer? I've looked over it several times and I don't see where the mistake might be. (I also edited; there was one significant error that I've corrected, where I forgot to actually antidifferentiate a term after setting it up to be easily antidifferentiated.) – Ian Jun 19 '15 at 14:25
  • @Ian: I wish I could but I don't have time now. I have already spent more time on this answer than I could afford. I need to literally go now (leave my house). – Rory Daulton Jun 19 '15 at 14:27