Solving Difficult Integral in Cosmology Lectures

[Onamor]

Homework Statement

Hi, this situtation arises in my cosmology lectures, but its purely mathematical:

I need to evaluate the LHS of \int \frac{da}{\sqrt{\frac{H_{0}^{2}\Omega_{0}}{a}+l^{-2}}}=dx^{0}-dx^{0}_{*}
using the substitution \frac{a}{l^{2}H^{2}_{0}\Omega_{0}}=\sin^{2}(\frac{u}{2})

Homework Equations

The answer is only possible/needed in parametric form:
a=\frac{l^{2}H^{2}_{0}\Omega_{0}}{2}(1-\cos(u))
dx^{0}-dx^{0}_{*}=\frac{l^{3}H^{2}_{0}\Omega_{0}}{2}(u-\sin(u))

The Attempt at a Solution

So first I differentiate the substitution \frac{da}{du}\frac{1}{l^{2}H^{2}_{0}\Omega_{0}}=\sin(\frac{u}{2})\cos(\frac{u}{2}) .
Then some algebra with the LHS: \int \frac{da}{\sqrt{\frac{H_{0}^{2}\Omega_{0}}{a}+l^{-2}}}=l\int \frac{da}{\sqrt{\frac{l^{2}H_{0}^{2}\Omega_{0}}{a}+1}}=l\int \frac{da}{\sqrt{\frac{1}{\sin^{2}(\frac{u}{2})}+1}}
=l\int \frac{da \sin(\frac{u}{2})}{\sqrt{1+sin^{2}(\frac{u}{2})}}}=l^{3}H^{2}_{0}\Omega_{0}\int \frac{du \sin^{2}(\frac{u}{2}) \cos(\frac{u}{2})}{\sqrt{1+sin^{2}(\frac{u}{2})}}}
Where in the last line I have used the substition. Giving me a very difficult integral which doesn't give the right answer anyway (according to Mathematica).

Any help is greatly appreciated

 

[diazona]

The substitution you're using does not convert 1/\sqrt{H_0^2\Omega_0/a + l^2} into l/\sqrt{l^2H_0^2\Omega_0/a + 1}. Check that mathematical step.

In any case, are you sure that's the substitution you should be using? I would think it should be a little different.

 

[Onamor]

AH! so sorry, I've misstyped the equation it should read
\int \frac{da}{\sqrt{\frac{H_{0}^{2}\Omega_{0}}{a}+l^{-2}}}=dx^{0}-dx^{0}_{*}
which is where my algabra comes from

So sorry, i will alter it in the OP now..

thank you very much for you're quick reply.
I've checked the substitution, and this is the form that my lecturer gives in his notes... but it did seem strange to me also.

 

[Onamor]

A side note on the choice of substitution:
If we had a minus infront of the l^{-2}, so;
\int \frac{da}{\sqrt{\frac{H_{0}^{2}\Omega_{0}}{a}-l^{-2}}}=dx^{0}-dx^{0}_{*}
then I would be suggested the substitution
\frac{a}{l^{2}H^{2}_{0}\Omega_{0}}=\sinh^{2}(\frac{u}{2})
and this would give the parametric solutions
a=\frac{l^{2}H^{2}_{0}\Omega_{0}}{2}(\cosh(u)-1)
dx^{0}-dx^{0}_{*}=\frac{l^{3}H^{2}_{0}\Omega_{0}}{2}(\sinh(u)-u)

 

[diazona]

Actually, it'd make more sense to use the hyperbolic sine for the plus case (the one you have), and the trigonometric sine for the minus case.

Bottom line, I don't see how the substitution given in your original post helps you solve the integral you have.