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# Very complicated integral

I'm looking for a solution to the following integral:

int (1/sqrt[1 - x^2 + a (x^6 - x^12)]) dx

where a is a parameter.

Could you help me to find a solution?

### 4 Answers by Expert Tutors

William S. | Experienced scientist, mathematician and instructor - WilliamExperienced scientist, mathematician and...
4.4 4.4 (10 lesson ratings) (10)
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If I'm interpreting this correctly, you are looking for the indefinite integral of

{sqrt[(1-x2 + a(x6-x12)]}-1 or [1-x2 + a(x6-x12]-1/2

Am I correct?

I look for a solution for this integral which contains a singularity for some points
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
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Try substitute x=sin(t); The integral becomes

∫dt/√[1+a*sin6(t)*(1+sin2(t)+sin4(t))];

This integral can be evaluated using the following expansion:

(1-x)=∑0(2k-1)!!/(2k)!!*xk, summed over k. This series converges for x<1.

Expression f(t)=a*sin6(t)(1+sin2(t)+sin4(t)) needs to be less than 1. This puts restrictions on the a parameter. [f(t)]k is integrable, however, integrals will be very ugly looking. For example,

∫f(t)dt=a*
213 x      337 Sin[2 x]        41 Sin[4 x]      31 Sin[6 x]
{-------  - ---------------     + -------------  -  -------------   +
256              12                    256                 1024

7 Sin[8 x]        Sin[10 x]
------------   -  ----------- }
2048                5120

As an addition to Kirill's method: the substitution x=sin(t) only works for |x|≤1. At x=±1, the integral becomes improper and needs to be split. For |x|≥1, you could use another trig substitution, perhaps x=sec(t).
In fact, I forgot a parameter in the formula, so the integral becomes:
int (1/sqrt[1 - x^2 + a (2(bx)^6 - (bx)^12)]) dx

I'm sorry for this error.
So now your potential is U(x) = a (2(bx)6-(bx)12). You should also mention that a and b are very small positive numbers. The equilibrium point now is at x0=1/b. Since x=1/r, b is the (stable) equilibrium distance between the interacting atoms/molecules, usually a few Angstroms. With U(1/b)=a, U'(1/b)=0, and U''(1/b)=-72ab, you get the quadratic approximation U(x)≈a(1-36 b(x-1/b)²) and can find the small frequency oscillations about the equilibrium.
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
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This integral represents the integral curves φ(r) (in polar coordinates, with x=1/r) for the famous 12-6 Lennard-Jones potential for atomic interactions. It is usually treated quantum-mechanically, not as a classical central potential problem.
You should graph the potential, U= a(x6-x12), and check it has a stable equilibrium at x0=2-1/6, corresponding to circular orbits.Deduce the nature of the other orbits (elliptic, parabolic, hyperbolic) qualitatively from your graph. Then do a Taylor series expansion of U about x0 up to order x2, so you will get a harmonic oscillator potential. Substitute this into your integral and get the well-know harmonic oscillator solutions. I doubt there exists an analytic expression for the unapproximated potential.