I want to show that for the metric:
$$ds^2=-X^2dT^2+dX^2$$
that,
$$\frac{dT}{d\tau}X^2=l, \quad l=constant$$
So $\frac{dT}{d\tau}X^2$ is a conserved quantity. So I know this can be done using killing vectors and things but I wanted to do it by using Christoffel symbols and the geodesic equation.
I calculated the Christoffel symbols and found that $\Gamma^{X}_{TX}=X,\Gamma^{T}_{TT}=X^3$ and $\Gamma^{T}_{TX}=X^3$, and all the others are $0$.
Which gives the following geodesic equations:
$$\frac{d^2T}{d\tau^2}+2X^3\frac{dT}{d\tau}\frac{dX}{d\tau}+X^4\bigg(\frac{dT}{d\tau}\bigg)^2=0\\ \frac{d^2X}{d\tau^2}+X\bigg(\frac{dT}{d\tau}\bigg)^2=0$$
I also have the normalization condition for massive particles $u^\mu u_\mu=u^\mu u^\nu g_{\mu\nu}=-1$ which implies that:
$$-X^2\bigg(\frac{dT}{d\tau}\bigg)^2+\bigg(\frac{dX}{d\tau}\bigg)^2=-1 $$
I can't seem to get the conserved quantity from this. I've tried starting from the beginning a few times to see where my mistake is but I'm really not sure where I'm going wrong, any ideas?
@Winther is correct the Christofell symbols are wrong they should be,
$$\Gamma^{T}_{TX}=1/X, \quad \Gamma^{X}_{TT}=X$$ and the rest are zero. This was a mistake with the inverse metric.
With these christofell symbols, the geodesic equation gives:
$$\frac{d^2T}{d\tau^2}+2/X\frac{dT}{d\tau}\frac{dX}{d\tau}=0\\ \frac{d^2X}{d\tau^2}+X\bigg(\frac{dT}{d\tau}\bigg)^2=0$$
Multiplying the first equation by $X^2$ gives
$$\frac{d^2T}{d\tau^2}X^2+2X\frac{dT}{d\tau}\frac{dX}{d\tau}=\frac{d}{d\tau}\Bigg(X^2\frac{dT}{d\tau}\Bigg)=0$$
Which directly integrates to give the conserved quantity.