Need help with an indefinate integral

[David_K]

  Hi,

  I came across this integral doing some research which I could not integrate.

May need to do some sort of substitution??   Any help will be appreciated. 

I have attached a microsoft word document with the integral.

Math_Help_Integral.docx

[studiot]

[math]\sqrt {\left( {A - B} \right){{\left( {\frac{k}{{r\cos \varphi }}} \right)}^2}} d\varphi  = \sqrt {\left( {A - B} \right)} \left( {\frac{k}{{r\cos \varphi }}} \right)d\varphi [/math]

 

[math] = C\frac{D}{{\cos \varphi }}d\varphi  = E\sec \varphi [/math]

 

Which is a standard integral.

[taeto]

Nice. But you should use \(\sqrt{x^2} = |x|\) instead of \(\sqrt{x^2} = x.\)

Edited August 1, 2018 by taeto

[David_K]

Hi,  Thanks for the quick reply. 

  I think the first expression was changed-- it's not (A - B)(k/cos x)²,  but A - B(k/cos x)²

 

 

 sorry, left out the r  -- should be (k/ r Cos x)²      -- Of course that makes little difference.

[taeto]

55 minutes ago, David_K said:

  I think the first expression was changed-- it's not (A - B)(k/cos x)²,  but A - B(k/cos x)²

 sorry, left out the r  -- should be (k/ r Cos x)²      -- Of course that makes little difference.

Do you assume anything about the values of A and B? E.g. if B is positive you have an imaginary integrand, so the question is if you are happy with a final answer that is not real? 

We can look at the very special case \( A = B\frac{k^2}{r^2}: \)

\[ \int \sqrt{ A - B(\frac{k}{r\cos x})^2 } dx = \int \sqrt{ A - A(1/\cos x)^2 } dx = \sqrt{A} \int \sqrt{1 - (1/\cos x)^2 } dx = \]

\[ \sqrt{A} \int \sqrt{\frac{\cos^2x - 1}{\cos^2x}} dx = \sqrt{A} \int \sqrt{- \frac{\sin^2x}{\cos^2x}} dx = \sqrt{-A} \int \sqrt{\tan^2x} dx = \sqrt{-A} \int | \tan x | dx. \]

The last indefinite integral does not appear standard. But its definite versions can calculated easily by splitting up into intervals with negative and positive values of the tangent function. So if I didn't make a silly mistake, this could be the most benign of the results you should expect.

Edited August 2, 2018 by taeto

[David_K]

 Astute observations Taeto...   We know that the expression B(k/ r Cos x)² is always at most =A,  and never negative, so that the radical is always real.   Does that help?

[taeto]

34 minutes ago, David_K said:

 Astute observations Taeto...   We know that the expression B(k/ r Cos x)² is always at most =A,  and never negative, so that the radical is always real.   Does that help?

So in that case you really only need the definite integral \( \int_0^t \) where \(0 \leq t < \pi/2. \) Because \( \cos x \) is symmetric around \(x =0,\) and the only way you can avoid an imaginary root is to stop some time before \( \pi/2, \) depending on the constants. 

By the trick of moving \(\sqrt{A}\) outside the integration, as in the "easy" special case, it seems enough to know the value of \( F_a(t) := \int_0^t \sqrt{1 - a\sec^2 x} dx \) for \(a \geq 0 \) and \(0 \leq t < \pi/2.\) It certainly seems to open it up nicely for a numerical approach. Whether it helps to get an analytical solution, I don't see that yet. 

Edited August 2, 2018 by taeto

[studiot]

Nice taeto, well done.

+1

[David_K]

  I agree,  Nice Taeto.  I was looking for an analytical solution though.    But nice to see the numerical approach.

[taeto]

I tried various substitutions, and checking with tables of known integrals. Unfortunately I see nothing to simplify things further, everything I try just seems to make it more ugly. It is not unlike my past history with potted plants.

I hope you can do something with it. Is it unfair to ask what kind of research this is related to? It looks a tiny bit cosmological, with the k and the r and the angles.

[David_K]

  Sorry to hear about your history of potted plants Taeto--lol.   And more sorry it wasn't a simple integral I had overlooked.  

 I will have to keep looking..  you could say it's kinda cosmological.

Thank you so much for your help... if anyone else can see a better solution, please let me know.

[David_K]

Hey all, it's been awhile on this post, but...

  I DID get the integral solution discussed above... for anyone interested, it is in the attachment:  (I'm not good at entering in lots of math in these threads)

 I readily admit I did NOT actually come upon this solution through all the hard work it would appear; I found a really cool integral website called

integral-calculator.com  <----  really good!!  if you need an integral.  They also have a related site for derivatives.

  I now have another question  I was hoping someone could help with:

  If we look at this integral solution, it clearly breaks down into 4 separate parts-- two of which are clearly imaginary.  and two which appear to have

an imaginary term  (that imaginary term can actually be totally eliminated, and those roots are real).

My question is this:  I am guessing these 4 separate parts of this solution might correspond to the 4 quadrants (where the trig function changes sign).

If the above guess is correct, is there any way to know which part of the solution corresponds to which quadrant it relates to... ???

 

   If you followed me this far, congratulations, and thanks for your effort.  Any ideas ??  

integral_solution03.docx