Free falling into a black hole: time to the singularity doubt!

I have trying to get a more precise insight into the calculation of the time to the singularity of a test mass into a non-rotatic uncharged black hole.

Prelude: the lagrangian for a massive particle in GR reads
begin{equation}
L=-msqrt{-g_{munu} dot{x}^mudot{ x}^nu}
end{equation}
Energy is
begin{equation}
-E=dfrac{partial L}{partial dot{t}}=-dfrac{m^2 g_{tt} dot{t}}{L}=constant
end{equation}
We also have
begin{equation}
g_{tt}dot{t}^2+g_{rr}dot{r}^2=-1
end{equation}
From this equation we get
begin{equation}
dot{r}^2=-left(1-dfrac{2GM}{r}right)+dfrac{E}{m}
end{equation}
Above, I used natural units with $c=1$ for simplicity. Taking $E=m$, leaves you with
begin{equation}
dot{r}^2=dfrac{2GM}{r}
end{equation}
that is indeed the classical escape velocity from a massive $M$ body for a test $m$ particle. Also, the dot means, I believe, differentiation with respect to proper or affine parameter $tau=lambda$.
However, I have seen two different procedures:

1st method. For a test mass $m$, the free fall time to the singularity can be estimated from the integral

begin{equation}
t=int dfrac{dr}{sqrt{-g_{rr}}}
end{equation}
That is, I integrate the negative of the radial “gravitational potential” (the metric field).
begin{equation}
t_s=frac{1}{c}int_0^{R_S}frac{1}{sqrt{dfrac{2GM}{c^2r}-1}}~dr=frac{pi GM}{c^3}=dfrac{pi R_S}{2c}simeq frac{M}{M_odot}times 1.54times 10^{-5}~{rm s}.
end{equation}
So, in summary,
begin{equation}
t_S = dfrac{pi GM}{c^3}
end{equation}

2nd method. In this method, we essentially integrate along a geodesic for the particle with mass m. For a particle mass with $E=m$, the free fall time to the singularity can be estimated from
begin{equation}
dfrac{dr}{dtau}=-sqrt{dfrac{2GM}{r}}
end{equation}
So, making a trip towards the singularity from the radial coordinate $r_0$ at $tau=0$, carries a time
begin{equation}
int_{r_0}^{r}r^{1/2}dr=-sqrt{2GM}int_0^tau dtau
end{equation}
begin{equation}
tau_S=dfrac{2}{3cR_S^{1/2}}left(r_0^{3/2}-r^{3/2}right)
end{equation}
Taking $r_0=R_S$ and $r=0$, you will obtain
begin{equation}
tau_S = dfrac{2}{3c}R_S= dfrac{4 GM}{3c^3}=dfrac{4}{3pi}t_S
end{equation}

I believe the second method is right, I think, but I saw the other method and I can not get its meaning, since I believe is "different" from the another one even when they are of the same order of magnitude.

Question: What is the meaning of these and how could this falling time methods be properly generalized to other black holes like Kerr black holes or SdS or RN?