# Integration problem ∫sqroot((cosx-cos^3(x))/(1-cos^3(x)))

• Krushnaraj Pandya
Problem statement 1. In summary, the problem is to find an elegant solution to a problem involving a trig function and an elliptic function of complex arguments.f

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## Homework Statement

Find
∫sqroot((cosx-cos^3(x))/(1-cos^3(x)))

2. The attempt at a solution
I tried taking cosx common on top which gives cosxsin^2(x), but it didn't give me any benefit, I tried splitting the numerator, factoring in sec^2(x) and a bunch of substitutions to no avail. Finally exhausted and out of insights I'm posting it here, would be glad if someone can help me with this

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Your parentheses mismatch. Are you taking square root of just numerator, or the whole thing?
Have you looked at manipulating the denominator? (a3 - b3)

Your parentheses mismatch. Are you taking square root of just numerator, or the whole thing?
Have you looked at manipulating the denominator? (a3 - b3)
The whole thing, sorry about that

The whole thing, sorry about that
I suggest you start using LaTeX. Here is your integral done that way:
$$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx.$$
To see how I typed that, just right-click on the image and ask for a display of math as tex commands.

scottdave
Im sure you noticed that you can take sin(x) outside of the sqrt.
But that doesn't seem to help either. It leads to one very messy solution.

Alternatively, you can factor out (1-cos(x)) from the numerator and denominator,
But that doesn't seem to help.

I think it's just plain ugly.

I suggest you start using LaTeX. Here is your integral done that way:
$$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx.$$
To see how I typed that, just right-click on the image and ask for a display of math as tex commands.
Thanks, I'll learn it immediately

Im sure you noticed that you can take sin(x) outside of the sqrt.
But that doesn't seem to help either. It leads to one very messy solution.

Alternatively, you can factor out (1-cos(x)) from the numerator and denominator,
But that doesn't seem to help.

I think it's just plain ugly.
hahaha, I know that...which is why I need the experts of physics forums on this

hahaha, I know that...which is why I need the experts of physics forums on this
If you're desparate, move the sin(x) outside of the sqrt and use www.wolframalpha.com.
But as I said, the result is not elegant.

If you're desparate, move the sin(x) outside of the sqrt and use www.wolframalpha.com.
But as I said, the result is not elegant.
I'm preparing for an UG entrance exam called JEE Advanced (for the best engineering institutes in India)...in a nutshell, they love asking these questions and there's surely an elegant solution which requires no use of technology whatsoever (a whole lot of mathematical aptitude though). The challenge is finding that trick

Your parentheses mismatch. Are you taking square root of just numerator, or the whole thing?
Have you looked at manipulating the denominator? (a3 - b3)
I'll try it and get back on it.
ok- factorized it, don't see anything useful still- now what?

I suggest you start using LaTeX. Here is your integral done that way:
$$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx.$$
To see how I typed that, just right-click on the image and ask for a display of math as tex commands.
I tried simply just copying those commands into the title, that's not how it works I guess. I'm having some trouble getting the hang of it, is there a proper tutorial?

I tried simply just copying those commands into the title, that's not how it works I guess. I'm having some trouble getting the hang of it, is there a proper tutorial?

It is usually not a good idea to have complicated formulas in the title; it would be better to say something like "integral with square root of trig functions", then state the problem details after the title (in section 1).

I'm preparing for an UG entrance exam called JEE Advanced (for the best engineering institutes in India)...in a nutshell, they love asking these questions and there's surely an elegant solution which requires no use of technology whatsoever (a whole lot of mathematical aptitude though). The challenge is finding that trick

If the problem really is as you have stated it, the indefinite integral involves Elliptic functions of complex arguments. It may simplify a bit, but the Elliptic functions will not go away. The integral really is "non-elementary".

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It is usually not a good idea to have complicated formulas in the title; it would be better to say something like "integral with square root of trig functions", then state the problem details after the title (in section 1).
ok, I'll keep that in mind

If the problem really is as you have stated it, the indefinite integral involves Elliptic functions of complex arguments. It may simplify a bit, but the Elliptic functions will not go away. The integral really is "non-elementary".
There is one way through which this gets easier, its still quite complex though. I did the following steps
1) took cos common, and then sin^2 out of the square root in numerator, giving sin in Num.
2)taking cosx=t, -sinxdx=dt (the sin in the numerator cancels out so now we only have cos terms which we can write in terms of t as -∫sqrt(t)/sqrt(1-t^3)
I wrote 1-t^3 as (1)^2-(t^(3/2))^2 (in perfect squares form), it looks a little bit like the derivative of cos inverse x, but I would have to put x as t^3/2 in its formula which is still a bit ugly and I'm hesitant to write it further, I'm quite sure cos inverse is involved though. (PS my textbook gives the answer as 2/3(Arccos(cos^(3/2)x)

Delta2
There is one way through which this gets easier, its still quite complex though. I did the following steps
1) took cos common, and then sin^2 out of the square root in numerator, giving sin in Num.
2)taking cosx=t, -sinxdx=dt (the sin in the numerator cancels out so now we only have cos terms which we can write in terms of t as -∫sqrt(t)/sqrt(1-t^3)
I wrote 1-t^3 as (1)^2-(t^(3/2))^2 (in perfect squares form), it looks a little bit like the derivative of cos inverse x, but I would have to put x as t^3/2 in its formula which is still a bit ugly and I'm hesitant to write it further, I'm quite sure cos inverse is involved though. (PS my textbook gives the answer as 2/3(Arccos(cos^(3/2)x)

I think you almost did it but you haven't realized it … Substituting ##u=t^{\frac{3}{2}}## then we have ##du=\frac{3}{2}\sqrt{t}dt##. At the numerator you already have ##\sqrt{t}dt##...

I think you almost did it but you haven't realized it … Substituting ##u=t^{\frac{3}{2}}## then we have ##du=\frac{3}{2}\sqrt{t}dt##. At the numerator you already have ##\sqrt{t}dt##...
Ah! yes, solved it! Thanks a lot