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?

please!

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?

please!

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?

please!

Tutors, sign in to answer this question.

If I'm interpreting this correctly, you are looking for the indefinite integral of

{sqrt[(1-x^{2} + a(x^{6}-x^{12})]}^{-1} or [1-x^{2} + a(x^{6}-x^{12}]^{-1/2}

Am I correct?

Try substitute x=sin(t); The integral becomes

∫dt/√[1+a*sin^{6}(t)*(1+sin^{2}(t)+sin^{4}(t))];

This integral can be evaluated using the following expansion:

(1-x)^{-½}=∑_{0}^{∞}(2k-1)!!/(2k)!!*x^{k}, summed over k. This series converges for x<1.

Expression f(t)=a*sin^{6}(t)(1+sin^{2}(t)+sin^{4}(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

{------- - --------------- + ------------- - ------------- +

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.

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 x_{0}=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.

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(x^{6}-x^{12}), and check it has a stable equilibrium at x_{0}=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 x_{0} up to order x^{2}, 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.

I think an analytical approach is more significant from the physical point of view, especially highlighting the phenomenons of orbiting and spiraling of the particles

If you find an analytic solution, you should publish it in JMC (Journal of Mathematical Chemistry)!

Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...

The Mathematica Integrator says that the integral of this function probably does not exist.

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