Show the integral below can be integrated to give Equation 1 θ(r) = ∫ {[(L/r^2) dr] / [ 2μ (E+(k/r) - (L^2/ 2μr^2))]^

Referring back to Eqs. (7), (8) and (9) of the solution of the previous problem (with appropriate changes in notation), you see that:

d theta/dt = L/(m r^2) (1)

dr/dt = Plus/minus [(2/m) * (E - V(r)) ]^(1/2) (2)

where V(r) = - k/r + L^2/(2m r^2) (3)

If you divide (1) by (2) you get:

d theta/dr = L/(m r^2) * [(2/m) * (E - V(r)) ]^(-1/2) =

L/r^2 * [ 2m (E - V(r)) ]^(-1/2) =

L/r^2 * [ 2m (E + k/r - L^2/(2m r^2) ) ]^(-1/2)

Here we've chosen the plus sign in (2), so that the particle moves in the direction of increasing theta (note that we start from r = minimal, so that r increases in the beginning).

We can integrate this to obtain:

theta = integral over r of L/r^2 * [ 2m (E + k/r - L^2/(2m r^2) ) ]^(-1/2) dr

Let's substitute r = L/U. Then dr = -L/U^2 dU and you obtain:

theta = - Integral over U of [ 2m (E + (k/L) U - U^2/(2m) ) ]^(-1/2) dU

The term in the square brackets can be written as:

T = m^2 K^2/L^2 + 2mE - (U - mk/L)^2

The integral of -1/sqrt[T] can be performed by noting that the integral of

-1/sqrt[a^2 - x^2] dx = arccos[x/a] + const.

So, the desired integral is:

theta = arccos[(U - mk/L)/a] + const

where:

a = sqrt[m^2 K^2/L^2 + 2mE ] (4)

If you write this as:

arccos[(U - mk/L)/a] = theta - const

and take cosine of both sides you get:

U - mk/L = a cos[theta - const]

If you put back U = L/r you get:

1/r = mk/L^2 + a/L cos[theta - const]

If we want theta=0 to correspond to the minimum distance, then the cos term should be maximal there and that implies that we can put the integration constant equal to zero. This means that the equation simplifies to:

1/r = mk/L^2 + a/L cos(theta)

Solving for cos(theta) and inserting the expression for a (Eq. (4)) gives the desired expression.